\displaystyle \textbf{Question 1: }~x^{1/x}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = x^{1/x}.

\displaystyle \text{Take the natural logarithm of both sides: } \ln y = \frac{1}{x}\ln x.

\displaystyle \text{Differentiate both sides with respect to } x: \frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\ln x}{x}\right).

\displaystyle \frac{d}{dx}\left(\frac{\ln x}{x}\right) = \frac{x\cdot\frac{1}{x} - \ln x\cdot 1}{x^2}.

\displaystyle \frac{1}{y}\frac{dy}{dx} = \frac{1 - \ln x}{x^2}.

\displaystyle \text{Multiply both sides by } y \text{ to isolate } \frac{dy}{dx}: \frac{dy}{dx} = y\cdot\frac{1 - \ln x}{x^2}.

\displaystyle \text{Substitute } y = x^{1/x} \text{ back into the expression: } \frac{dy}{dx} = x^{1/x}\cdot\frac{1 - \ln x}{x^2}.

\displaystyle \textbf{Question 2: }~x^{\sin x}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = x^{\sin x} \quad \text{...(i)}.

\displaystyle \text{Taking natural logarithm on both sides,} \ \ln y = \ln\left(x^{\sin x}\right).

\displaystyle \Rightarrow \ln y = \sin x \, \ln x \quad [\because \ln(a^b) = b\ln a].

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \sin x \frac{d}{dx}(\ln x) + \ln x \frac{d}{dx}(\sin x) \quad [\text{using product rule}].

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \sin x\left(\frac{1}{x}\right) + (\ln x)(\cos x).

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[\frac{\sin x}{x} + (\ln x)(\cos x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = x^{\sin x}\left[\frac{\sin x}{x} + (\ln x)(\cos x)\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 3: }~(1+\cos x)^x
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = (1+\cos x)^x \quad \text{...(i)}.

\displaystyle \text{Taking natural logarithm on both sides, } \ln y = \ln\!\left((1+\cos x)^x\right).

\displaystyle \Rightarrow \ln y = x \ln(1+\cos x).

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\,\frac{d}{dx}\bigl[\ln(1+\cos x)\bigr] + \ln(1+\cos x)\,\frac{d}{dx}(x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = x \cdot \frac{1}{1+\cos x}\cdot(-\sin x) + \ln(1+\cos x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \ln(1+\cos x) - \frac{x\sin x}{1+\cos x}.

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[\ln(1+\cos x) - \frac{x\sin x}{1+\cos x}\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (1+\cos x)^x \left[\ln(1+\cos x) - \frac{x\sin x}{1+\cos x}\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 4: }~x^{\cos^{-1}x}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = x^{\cos^{-1} x} \quad \text{...(i)}.

\displaystyle \text{Taking natural logarithm on both sides, } \ln y = \ln\!\left(x^{\cos^{-1} x}\right).

\displaystyle \Rightarrow \ln y = \cos^{-1} x \, \ln x.

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \cos^{-1} x \frac{d}{dx}(\ln x) + \ln x \frac{d}{dx}(\cos^{-1} x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \cos^{-1} x \left(\frac{1}{x}\right) + \ln x \left(-\frac{1}{\sqrt{1-x^2}}\right).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{\cos^{-1} x}{x} - \frac{\ln x}{\sqrt{1-x^2}}.

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[\frac{\cos^{-1} x}{x} - \frac{\ln x}{\sqrt{1-x^2}}\right].

\displaystyle \Rightarrow \frac{dy}{dx} = x^{\cos^{-1} x} \left[\frac{\cos^{-1} x}{x} - \frac{\ln x}{\sqrt{1-x^2}}\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 5: }~(\log x)^x
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = (\log x)^x \quad \text{...(i)}.

\displaystyle \text{Taking logarithm on both sides, } \log y = \log\!\left((\log x)^x\right).

\displaystyle \Rightarrow \log y = x \log(\log x).

\displaystyle \text{Differentiating both sides with respect to } x \text{ (using product rule),}

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\,\frac{d}{dx}\bigl[\log(\log x)\bigr] + \log(\log x)\,\frac{d}{dx}(x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\left(\frac{1}{\log x}\cdot\frac{1}{x}\right) + \log(\log x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{1}{\log x} + \log(\log x).

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[\frac{1}{\log x} + \log(\log x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (\log x)^x \left[\frac{1}{\log x} + \log(\log x)\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 6: }~(\log x)^{\cos x}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = (\log x)^{\cos x} \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log y = \log\!\left((\log x)^{\cos x}\right).

\displaystyle \Rightarrow \log y = \cos x \, \log(\log x).

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \cos x \frac{d}{dx}\bigl[\log(\log x)\bigr] + \log(\log x)\frac{d}{dx}(\cos x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \cos x \left(\frac{1}{\log x}\cdot\frac{1}{x}\right) + \log(\log x)(-\sin x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{\cos x}{x\log x} - \sin x \log(\log x).

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[\frac{\cos x}{x\log x} - \sin x \log(\log x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (\log x)^{\cos x} \left[\frac{\cos x}{x\log x} - \sin x \log(\log x)\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 7: }~(\sin x)^{\cos x}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = (\sin x)^{\cos x} \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log y = \log\!\left((\sin x)^{\cos x}\right).

\displaystyle \Rightarrow \log y = \cos x \, \log(\sin x).

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \cos x \frac{d}{dx}\bigl[\log(\sin x)\bigr] + \log(\sin x)\frac{d}{dx}(\cos x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \cos x \left(\frac{1}{\sin x}\cdot\cos x\right) + \log(\sin x)(-\sin x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \cos x \cot x - \sin x \log(\sin x).

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[\cos x \cot x - \sin x \log(\sin x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (\sin x)^{\cos x} \left[\cos x \cot x - \sin x \log(\sin x)\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 8: }~e^{x\log x}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = e^{x\log x}.

\displaystyle \text{Taking log on both sides, } \log y = \log\!\left(e^{x\log x}\right).

\displaystyle \Rightarrow \log y = x\log x \cdot \log e.

\displaystyle \Rightarrow \log y = x\log x \quad (\because \log e = 1).

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\frac{d}{dx}(\log x) + \log x \frac{d}{dx}(x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\left(\frac{1}{x}\right) + \log x(1).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = 1 + \log x.

\displaystyle \Rightarrow \frac{dy}{dx} = y(1+\log x).

\displaystyle \Rightarrow \frac{dy}{dx} = e^{x\log x}(1+\log x).

\displaystyle \Rightarrow \frac{dy}{dx} = e^{\log x^x}(1+\log x).

\displaystyle \Rightarrow \frac{dy}{dx} = x^x(1+\log x).

\displaystyle \textbf{Question 9: }~(\sin x)^{\log x}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = (\sin x)^{\log x} \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log y = \log\!\left((\sin x)^{\log x}\right).

\displaystyle \Rightarrow \log y = \log x \, \log(\sin x).

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log x \frac{d}{dx}\bigl[\log(\sin x)\bigr] + \log(\sin x)\frac{d}{dx}(\log x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log x \left(\frac{1}{\sin x}\cdot\cos x\right) + \log(\sin x)\left(\frac{1}{x}\right).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log x \cot x + \frac{\log(\sin x)}{x}.

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[\log x \cot x + \frac{\log(\sin x)}{x}\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (\sin x)^{\log x} \left[\log x \cot x + \frac{\log(\sin x)}{x}\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 10: }~10^{\log(\sin x)}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = 10^{\log(\sin x)} \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log y = \log\!\left(10^{\log(\sin x)}\right).

\displaystyle \Rightarrow \log y = \log(\sin x)\,\log 10.

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log 10 \,\frac{d}{dx}\bigl[\log(\sin x)\bigr].

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log 10 \left(\frac{1}{\sin x}\cdot\cos x\right).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log 10 \cdot \cot x.

\displaystyle \Rightarrow \frac{dy}{dx} = y\,[\log 10 \cdot \cot x].

\displaystyle \Rightarrow \frac{dy}{dx} = 10^{\log(\sin x)} \cdot \log 10 \cdot \cot x \quad [\text{From (i)}].

\displaystyle \textbf{Question 11: }~(\log x)^{\log x}\;
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = (\log x)^{\log x} \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log y = \log\!\left((\log x)^{\log x}\right).

\displaystyle \Rightarrow \log y = \log x \cdot \log(\log x).

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log(\log x)\frac{d}{dx}(\log x) + \log x \frac{d}{dx}\bigl[\log(\log x)\bigr].

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log(\log x)\left(\frac{1}{x}\right) + \log x \left(\frac{1}{\log x}\cdot\frac{1}{x}\right).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{\log(\log x)}{x} + \frac{1}{x}.

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[\frac{1+\log(\log x)}{x}\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (\log x)^{\log x} \left[\frac{1+\log(\log x)}{x}\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 12: }~10^{(10^x)}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = 10^{10^x} \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log y = \log\!\left(10^{10^x}\right).

\displaystyle \Rightarrow \log y = 10^x \log 10.

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log 10 \cdot \frac{d}{dx}(10^x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log 10 \cdot 10^x \log 10.

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = 10^x (\log 10)^2.

\displaystyle \Rightarrow \frac{dy}{dx} = y\,[10^x (\log 10)^2].

\displaystyle \Rightarrow \frac{dy}{dx} = 10^{10^x} \cdot 10^x (\log 10)^2 \quad [\text{From (i)}].

\displaystyle \textbf{Question 13: }~\sin\left(x^x\right)
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = \sin(x^x).

\displaystyle \Rightarrow \sin^{-1} y = x^x \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log(\sin^{-1} y) = \log(x^x).

\displaystyle \Rightarrow \log(\sin^{-1} y) = x \log x.

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{\sin^{-1} y}\frac{d}{dx}(\sin^{-1} y) = x\frac{d}{dx}(\log x) + \log x \frac{d}{dx}(x).

\displaystyle \Rightarrow \frac{1}{\sin^{-1} y} \left(\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}\right) = x\left(\frac{1}{x}\right) + \log x.

\displaystyle \Rightarrow \frac{1}{\sin^{-1} y} \left(\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}\right) = 1 + \log x.

\displaystyle \Rightarrow \frac{dy}{dx} = \sin^{-1} y \sqrt{1-y^2}\,(1+\log x).

\displaystyle \Rightarrow \frac{dy}{dx} = \sin^{-1}(\sin x^x)\sqrt{1-(\sin x^x)^2}\,(1+\log x).

\displaystyle \Rightarrow \frac{dy}{dx} = x^x \cos x^x \,(1+\log x) \quad [\text{using equation (i)}].

\displaystyle \textbf{Question 14: }~(\sin^{-1}x)^x
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = (\sin^{-1} x)^x \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log y = \log\!\left((\sin^{-1} x)^x\right).

\displaystyle \Rightarrow \log y = x \log(\sin^{-1} x).

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\frac{d}{dx}\bigl[\log(\sin^{-1} x)\bigr] + \log(\sin^{-1} x)\frac{d}{dx}(x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\left(\frac{1}{\sin^{-1} x}\cdot\frac{1}{\sqrt{1-x^2}}\right) + \log(\sin^{-1} x).

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[\log(\sin^{-1} x) + \frac{x}{\sin^{-1} x\sqrt{1-x^2}}\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (\sin^{-1} x)^x \left[\log(\sin^{-1} x) + \frac{x}{\sin^{-1} x\sqrt{1-x^2}}\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 15: }~x^{\sin^{-1}x}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = x^{\sin^{-1} x} \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log y = \log\!\left(x^{\sin^{-1} x}\right).

\displaystyle \Rightarrow \log y = \sin^{-1} x \, \log x.

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \sin^{-1} x \frac{d}{dx}(\log x) + (\log x)\frac{d}{dx}(\sin^{-1} x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \sin^{-1} x\left(\frac{1}{x}\right) + (\log x)\left(\frac{1}{\sqrt{1-x^2}}\right).

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[\frac{\sin^{-1} x}{x} + \frac{\log x}{\sqrt{1-x^2}}\right].

\displaystyle \Rightarrow \frac{dy}{dx} = x^{\sin^{-1} x} \left[\frac{\sin^{-1} x}{x} + \frac{\log x}{\sqrt{1-x^2}}\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 16: }~(\tan x)^{1/x}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = (\tan x)^{1/x} \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log y = \log\!\left((\tan x)^{1/x}\right).

\displaystyle \Rightarrow \log y = \frac{1}{x}\log(\tan x).

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{1}{x}\frac{d}{dx}\bigl[\log(\tan x)\bigr] + \log(\tan x)\frac{d}{dx}\left(\frac{1}{x}\right).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{1}{x}\left(\frac{1}{\tan x}\cdot\sec^2 x\right) + \log(\tan x)\left(-\frac{1}{x^2}\right).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{\sec^2 x}{x\tan x} - \frac{\log(\tan x)}{x^2}.

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[\frac{\sec^2 x}{x\tan x} - \frac{\log(\tan x)}{x^2}\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (\tan x)^{1/x} \left[\frac{\sec^2 x}{x\tan x} - \frac{\log(\tan x)}{x^2}\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 17: }~x^{\tan^{-1}x}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = x^{\tan^{-1} x} \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log y = \log\!\left(x^{\tan^{-1} x}\right).

\displaystyle \Rightarrow \log y = \tan^{-1} x \, \log x.

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \tan^{-1} x \frac{d}{dx}(\log x) + \log x \frac{d}{dx}(\tan^{-1} x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \tan^{-1} x\left(\frac{1}{x}\right) + \log x\left(\frac{1}{1+x^2}\right).

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[\frac{\tan^{-1} x}{x} + \frac{\log x}{1+x^2}\right].

\displaystyle \Rightarrow \frac{dy}{dx} = x^{\tan^{-1} x} \left[\frac{\tan^{-1} x}{x} + \frac{\log x}{1+x^2}\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 18: }

\displaystyle \textbf{(i): }~(x^x)\sqrt{x}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = x^x\sqrt{x} \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log y = \log(x^x\sqrt{x}).

\displaystyle \Rightarrow \log y = \log(x^x) + \log(x^{1/2}).

\displaystyle \Rightarrow \log y = x\log x + \frac{1}{2}\log x.

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\frac{d}{dx}(\log x) + \log x\frac{d}{dx}(x) + \frac{1}{2}\frac{d}{dx}(\log x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\left(\frac{1}{x}\right) + \log x(1) + \frac{1}{2}\left(\frac{1}{x}\right).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = 1 + \log x + \frac{1}{2x}.

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[1 + \log x + \frac{1}{2x}\right].

\displaystyle \Rightarrow \frac{dy}{dx} = x^x\sqrt{x}\left[1 + \log x + \frac{1}{2x}\right] \quad [\text{From (i)}].

\displaystyle \Rightarrow \frac{dy}{dx} = x^{x+\frac{1}{2}}\left[\log x + \frac{2x+1}{2x}\right].

\displaystyle \textbf{(ii): }~x^{(\sin x-\cos x)}+\frac{x^2-1}{x^2+1}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = x^{\sin x-\cos x} + \frac{x^2-1}{x^2+1}.

\displaystyle \Rightarrow y = e^{\log x^{(\sin x-\cos x)}} + \frac{x^2-1}{x^2+1}.

\displaystyle \Rightarrow y = e^{(\sin x-\cos x)\log x} + \frac{x^2-1}{x^2+1}.

\displaystyle \text{Differentiate with respect to } x \text{ using chain rule,}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{d}{dx}\!\left[e^{(\sin x-\cos x)\log x}\right] + \frac{d}{dx}\!\left(\frac{x^2-1}{x^2+1}\right).

\displaystyle \Rightarrow \frac{dy}{dx} = e^{(\sin x-\cos x)\log x} \frac{d}{dx}\!\left[(\sin x-\cos x)\log x\right] + \frac{(x^2+1)\,2x-(x^2-1)\,2x}{(x^2+1)^2}.

\displaystyle \Rightarrow \frac{dy}{dx} = e^{(\sin x-\cos x)\log x} \left[(\sin x-\cos x)\frac{1}{x} + \log x(\sin x+\cos x)\right] + \frac{4x}{(x^2+1)^2}.

\displaystyle \Rightarrow \frac{dy}{dx} = x^{\sin x-\cos x} \left[\frac{\sin x-\cos x}{x} + (\sin x+\cos x)\log x\right] + \frac{4x}{(x^2+1)^2}.

\displaystyle \textbf{(iii): }~x^x\cos x+\frac{x^2+1}{x^2-1}\;[\text{CBSE 2011}]
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = x^{x\cos x} + \frac{x^2+1}{x^2-1}.

\displaystyle \text{Also, let } u = x^{x\cos x} \text{ and } v = \frac{x^2+1}{x^2-1}.

\displaystyle \therefore y = u + v.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \quad \text{...(i)}.

\displaystyle \text{Now, } u = x^{x\cos x}.

\displaystyle \Rightarrow \log u = \log(x^{x\cos x}).

\displaystyle \Rightarrow \log u = x\cos x \log x.

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = \cos x \log x \frac{d}{dx}(x) + x \log x \frac{d}{dx}(\cos x) + x \cos x \frac{d}{dx}(\log x).

\displaystyle \Rightarrow \frac{du}{dx} = u\left[\cos x \log x + x(-\sin x)\log x + x\cos x\left(\frac{1}{x}\right)\right].

\displaystyle \Rightarrow \frac{du}{dx} = x^{x\cos x}\left[\cos x(1+\log x) - x\sin x \log x\right] \quad \text{...(ii)}.

\displaystyle \text{Again, } v = \frac{x^2+1}{x^2-1}.

\displaystyle \Rightarrow \log v = \log(x^2+1) - \log(x^2-1).

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} = \frac{2x}{x^2+1} - \frac{2x}{x^2-1}.

\displaystyle \Rightarrow \frac{dv}{dx} = v\left[\frac{2x(x^2-1)-2x(x^2+1)}{(x^2+1)(x^2-1)}\right].

\displaystyle \Rightarrow \frac{dv}{dx} = \frac{x^2+1}{x^2-1}\left[\frac{-4x}{(x^2+1)(x^2-1)}\right].

\displaystyle \Rightarrow \frac{dv}{dx} = -\frac{4x}{(x^2-1)^2} \quad \text{...(iii)}.

\displaystyle \text{From (i), (ii) and (iii), we obtain}

\displaystyle \frac{dy}{dx} = x^{x\cos x}\left[\cos x(1+\log x) - x\sin x \log x\right] - \frac{4x}{(x^2-1)^2}.

\displaystyle \textbf{(iv): }~(x\cos x)^x+(x\sin x)^{1/x}\;
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = (x\cos x)^x + (x\sin x)^{1/x}.

\displaystyle \text{Also, let } u = (x\cos x)^x \text{ and } v = (x\sin x)^{1/x}.

\displaystyle \therefore y = u + v.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \quad \text{...(i)}.

\displaystyle \text{Now, } u = (x\cos x)^x.

\displaystyle \Rightarrow \log u = \log\!\left((x\cos x)^x\right).

\displaystyle \Rightarrow \log u = x\log(x\cos x).

\displaystyle \Rightarrow \log u = x[\log x + \log(\cos x)].

\displaystyle \Rightarrow \log u = x\log x + x\log(\cos x).

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = \frac{d}{dx}(x\log x) + \frac{d}{dx}[x\log(\cos x)].

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = [\log x + 1] + [\log(\cos x) - x\tan x].

\displaystyle \Rightarrow \frac{du}{dx} = (x\cos x)^x\,[1 - x\tan x + \log(x\cos x)] \quad \text{...(ii)}.

\displaystyle \text{Again, } v = (x\sin x)^{1/x}.

\displaystyle \Rightarrow \log v = \log\!\left((x\sin x)^{1/x}\right).

\displaystyle \Rightarrow \log v = \frac{1}{x}\log(x\sin x).

\displaystyle \Rightarrow \log v = \frac{1}{x}[\log x + \log(\sin x)].

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} = \frac{d}{dx}\!\left(\frac{1}{x}\log x\right) + \frac{d}{dx}\!\left(\frac{1}{x}\log(\sin x)\right).

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} = \frac{1-\log x}{x^2} + \frac{-\log(\sin x) + x\cot x}{x^2}.

\displaystyle \Rightarrow \frac{dv}{dx} = (x\sin x)^{1/x} \left[\frac{1-\log(x\sin x)+x\cot x}{x^2}\right] \quad \text{...(iii)}.

\displaystyle \text{From (i), (ii) and (iii), we obtain}

\displaystyle \frac{dy}{dx} = (x\cos x)^x[1 - x\tan x + \log(x\cos x)] + (x\sin x)^{1/x} \left[\frac{x\cot x + 1 - \log(x\sin x)}{x^2}\right].

\displaystyle \textbf{(v): }~\left(x+\frac{1}{x}\right)^x+x\left(1+\frac{1}{x}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = \left(x+\frac{1}{x}\right)^x + x^{\left(1+\frac{1}{x}\right)}.

\displaystyle \text{Also, let } u = \left(x+\frac{1}{x}\right)^x \text{ and } v = x^{\left(1+\frac{1}{x}\right)}.

\displaystyle \therefore y = u + v.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \quad \text{...(i)}.

\displaystyle \text{Now, } u = \left(x+\frac{1}{x}\right)^x.

\displaystyle \Rightarrow \log u = \log\!\left[\left(x+\frac{1}{x}\right)^x\right].

\displaystyle \Rightarrow \log u = x\log\!\left(x+\frac{1}{x}\right).

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = \log\!\left(x+\frac{1}{x}\right) + x\frac{1}{\left(x+\frac{1}{x}\right)}\frac{d}{dx}\!\left(x+\frac{1}{x}\right).

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = \log\!\left(x+\frac{1}{x}\right) + \frac{x}{\left(x+\frac{1}{x}\right)}\left(1-\frac{1}{x^2}\right).

\displaystyle \Rightarrow \frac{du}{dx} = \left(x+\frac{1}{x}\right)^x \left[\log\!\left(x+\frac{1}{x}\right) + \frac{x^2-1}{x^2+1}\right] \quad \text{...(ii)}.

\displaystyle \text{Again, } v = x^{\left(1+\frac{1}{x}\right)}.

\displaystyle \Rightarrow \log v = \log\!\left[x^{\left(1+\frac{1}{x}\right)}\right].

\displaystyle \Rightarrow \log v = \left(1+\frac{1}{x}\right)\log x.

\displaystyle \text{Differentiating both sides with respect to } x,

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} = \log x \frac{d}{dx}\!\left(1+\frac{1}{x}\right) + \left(1+\frac{1}{x}\right)\frac{d}{dx}(\log x).

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} = -\frac{\log x}{x^2} + \frac{1}{x} + \frac{1}{x^2}.

\displaystyle \Rightarrow \frac{dv}{dx} = x^{\left(1+\frac{1}{x}\right)} \left(\frac{x+1-\log x}{x^2}\right) \quad \text{...(iii)}.

\displaystyle \text{From (i), (ii) and (iii), we obtain}

\displaystyle \frac{dy}{dx} = \left(x+\frac{1}{x}\right)^x \left[\frac{x^2-1}{x^2+1} + \log\!\left(x+\frac{1}{x}\right)\right] + x^{\left(1+\frac{1}{x}\right)} \left(\frac{x+1-\log x}{x^2}\right).

\displaystyle \textbf{(vi): }~e^{\sin x}+(\tan x)^x\;[\text{CBSE 2003}]
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = e^{\sin x} + (\tan x)^x.

\displaystyle \Rightarrow y = e^{\sin x} + e^{x\log(\tan x)}.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{d}{dx}\!\left(e^{\sin x}\right) + \frac{d}{dx}\!\left(e^{x\log(\tan x)}\right).

\displaystyle \Rightarrow \frac{dy}{dx} = e^{\sin x}\frac{d}{dx}(\sin x) + e^{x\log(\tan x)}\frac{d}{dx}\!\left[x\log(\tan x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = e^{\sin x}(\cos x) + e^{x\log(\tan x)} \left[x\frac{d}{dx}(\log(\tan x)) + \log(\tan x)\frac{d}{dx}(x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = e^{\sin x}(\cos x) + (\tan x)^x \left[x\left(\frac{1}{\tan x}\sec^2 x\right) + \log(\tan x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = e^{\sin x}(\cos x) + (\tan x)^x \left[x\sec x\,\mathrm{cosec} x + \log(\tan x)\right].

\displaystyle \textbf{(vii): }~(\cos x)^x+(\sin x)^{1/x}\;[\text{CBSE 2010}]
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = (\cos x)^x + (\sin x)^{1/x}.

\displaystyle \Rightarrow y = e^{x\log(\cos x)} + e^{\frac{1}{x}\log(\sin x)}.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{d}{dx}\!\left(e^{x\log(\cos x)}\right) + \frac{d}{dx}\!\left(e^{\frac{1}{x}\log(\sin x)}\right).

\displaystyle \Rightarrow \frac{dy}{dx} = e^{x\log(\cos x)}\frac{d}{dx}\!\left[x\log(\cos x)\right] + e^{\frac{1}{x}\log(\sin x)}\frac{d}{dx}\!\left(\frac{1}{x}\log(\sin x)\right).

\displaystyle \Rightarrow \frac{dy}{dx} = e^{x\log(\cos x)} \left[x\frac{d}{dx}(\log(\cos x)) + \log(\cos x)\frac{d}{dx}(x)\right] + e^{\frac{1}{x}\log(\sin x)} \left[\frac{1}{x}\frac{d}{dx}(\log(\sin x)) + \log(\sin x)\frac{d}{dx}\!\left(\frac{1}{x}\right)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (\cos x)^x\,[x(-\tan x)+\log(\cos x)] + (\sin x)^{1/x} \left[\frac{1}{x}\cot x - \frac{1}{x^2}\log(\sin x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (\cos x)^x\,[\log(\cos x)-x\tan x] + (\sin x)^{1/x} \left[\frac{\cot x}{x}-\frac{\log(\sin x)}{x^2}\right].

\displaystyle \textbf{(viii): }~x^{x^2-3}+(x-3)^{x^2}\;
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = x^{x^2-3} + (x-3)^{x^2}.

\displaystyle \text{Also, let } u = x^{x^2-3} \text{ and } v = (x-3)^{x^2}.

\displaystyle \therefore y = u + v.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \quad \text{...(i)}.

\displaystyle \text{Now, } u = x^{x^2-3}.

\displaystyle \Rightarrow \log u = \log\!\left(x^{x^2-3}\right).

\displaystyle \Rightarrow \log u = (x^2-3)\log x.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = \log x\frac{d}{dx}(x^2-3) + (x^2-3)\frac{d}{dx}(\log x).

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = 2x\log x + \frac{x^2-3}{x}.

\displaystyle \Rightarrow \frac{du}{dx} = x^{x^2-3}\left[\frac{x^2-3}{x} + 2x\log x\right].

\displaystyle \text{Again, } v = (x-3)^{x^2}.

\displaystyle \Rightarrow \log v = \log\!\left((x-3)^{x^2}\right).

\displaystyle \Rightarrow \log v = x^2\log(x-3).

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} = \log(x-3)\frac{d}{dx}(x^2) + x^2\frac{d}{dx}[\log(x-3)].

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} = 2x\log(x-3) + \frac{x^2}{x-3}.

\displaystyle \Rightarrow \frac{dv}{dx} = (x-3)^{x^2}\left[\frac{x^2}{x-3} + 2x\log(x-3)\right].

\displaystyle \text{Substituting in equation (i), we obtain}

\displaystyle \frac{dy}{dx} = x^{x^2-3}\left[\frac{x^2-3}{x} + 2x\log x\right] + (x-3)^{x^2}\left[\frac{x^2}{x-3} + 2x\log(x-3)\right].

\displaystyle \textbf{Question 19: }~y=e^x+10^x+x^x
\displaystyle \text{Answer:}

\displaystyle \text{We have } y = e^x + 10^x + x^x.

\displaystyle \Rightarrow y = e^x + 10^x + e^{\log x^x}.

\displaystyle \Rightarrow y = e^x + 10^x + e^{x\log x}.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{d}{dx}(e^x) + \frac{d}{dx}(10^x) + \frac{d}{dx}\!\left(e^{x\log x}\right).

\displaystyle \Rightarrow \frac{dy}{dx} = e^x + 10^x\log 10 + e^{x\log x}\frac{d}{dx}(x\log x).

\displaystyle \Rightarrow \frac{dy}{dx} = e^x + 10^x\log 10 + e^{x\log x} \left[x\frac{d}{dx}(\log x) + \log x\frac{d}{dx}(x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = e^x + 10^x\log 10 + e^{x\log x}\left[1+\log x\right].

\displaystyle \Rightarrow \frac{dy}{dx} = e^x + 10^x\log 10 + x^x(1+\log x).

\displaystyle \textbf{Question 20: }~y=x^n+n^x+x^x+n^n
\displaystyle \text{Answer:}

\displaystyle \text{We have } y = x^n + n^x + x^x + n^n, \text{ where } n \text{ is a constant}.

\displaystyle \Rightarrow y = x^n + n^x + e^{x\log x} + n^n.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{d}{dx}(x^n) + \frac{d}{dx}(n^x) + \frac{d}{dx}\!\left(e^{x\log x}\right) + \frac{d}{dx}(n^n).

\displaystyle \Rightarrow \frac{dy}{dx} = nx^{n-1} + n^x\log n + e^{x\log x}\frac{d}{dx}(x\log x) + 0.

\displaystyle \Rightarrow \frac{dy}{dx} = nx^{n-1} + n^x\log n + e^{x\log x}\left[x\frac{d}{dx}(\log x) + \log x\frac{d}{dx}(x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = nx^{n-1} + n^x\log n + e^{x\log x}(1+\log x).

\displaystyle \Rightarrow \frac{dy}{dx} = nx^{n-1} + n^x\log n + x^x(1+\log x).

\displaystyle \textbf{Question 21: }~y=\dfrac{(x^2-1)^3(2x-1)}{\sqrt{(x-3)(4x-1)}}
\displaystyle \text{Answer:}

\displaystyle \text{We have } y = \frac{(x^2-1)^3(2x-1)}{\sqrt{(x-3)(4x-1)}} \quad \text{...(i)}.

\displaystyle \Rightarrow y = \frac{(x^2-1)^3(2x-1)}{(x-3)^{1/2}(4x-1)^{1/2}}.

\displaystyle \text{Taking log on both sides, }

\displaystyle \log y = \log\!\left[\frac{(x^2-1)^3(2x-1)}{(x-3)^{1/2}(4x-1)^{1/2}}\right].

\displaystyle \Rightarrow \log y = \log(x^2-1)^3 + \log(2x-1) - \log(x-3)^{1/2} - \log(4x-1)^{1/2}.

\displaystyle \Rightarrow \log y = 3\log(x^2-1) + \log(2x-1) - \frac{1}{2}\log(x-3) - \frac{1}{2}\log(4x-1).

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = 3\frac{d}{dx}[\log(x^2-1)] + \frac{d}{dx}[\log(2x-1)] - \frac{1}{2}\frac{d}{dx}[\log(x-3)] - \frac{1}{2}\frac{d}{dx}[\log(4x-1)].

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = 3\left(\frac{1}{x^2-1}\right)(2x) + \frac{1}{2x-1}(2) - \frac{1}{2}\left(\frac{1}{x-3}\right)(1) - \frac{1}{2}\left(\frac{1}{4x-1}\right)(4).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{6x}{x^2-1} + \frac{2}{2x-1} - \frac{1}{2(x-3)} - \frac{2}{4x-1}.

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[\frac{6x}{x^2-1} + \frac{2}{2x-1} - \frac{1}{2(x-3)} - \frac{2}{4x-1}\right].

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{(x^2-1)^3(2x-1)}{\sqrt{(x-3)(4x-1)}} \left[\frac{6x}{x^2-1} + \frac{2}{2x-1} - \frac{1}{2(x-3)} - \frac{2}{4x-1}\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 22: }~y=\dfrac{e^{ax}\sec x\log x}{\sqrt{1-2x}}
\displaystyle \text{Answer:}

\displaystyle \text{We have } y = \frac{e^{ax}\sec x \log x}{\sqrt{1-2x}} \quad \text{...(i)}.

\displaystyle \Rightarrow y = \frac{e^{ax}\sec x \log x}{(1-2x)^{1/2}}.

\displaystyle \text{Taking log on both sides,}

\displaystyle \log y = \log(e^{ax}) + \log(\sec x) + \log(\log x) - \frac{1}{2}\log(1-2x).

\displaystyle \Rightarrow \log y = ax + \log(\sec x) + \log(\log x) - \frac{1}{2}\log(1-2x).

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}(ax) + \frac{d}{dx}[\log(\sec x)] + \frac{d}{dx}[\log(\log x)] - \frac{1}{2}\frac{d}{dx}[\log(1-2x)].

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = a + \tan x + \frac{1}{x\log x} + \frac{1}{1-2x}.

\displaystyle \Rightarrow \frac{dy}{dx} = y\left[a + \tan x + \frac{1}{x\log x} + \frac{1}{1-2x}\right].

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{e^{ax}\sec x \log x}{\sqrt{1-2x}} \left[a + \tan x + \frac{1}{x\log x} + \frac{1}{1-2x}\right] \quad [\text{From (i)}].

\displaystyle \textbf{Question 23: }~y=e^{3x}\sin 4x\cdot 2^x
\displaystyle \text{Answer:}

\displaystyle \text{We have } y = e^{3x}\,\sin 4x \, 2^x \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log y = \log(e^{3x}) + \log(\sin 4x) + \log(2^x).

\displaystyle \Rightarrow \log y = 3x\log e + \log(\sin 4x) + x\log 2.

\displaystyle \Rightarrow \log y = 3x + \log(\sin 4x) + x\log 2.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}(3x) + \frac{d}{dx}[\log(\sin 4x)] + \frac{d}{dx}(x\log 2).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = 3 + \frac{1}{\sin 4x}\frac{d}{dx}(\sin 4x) + \log 2.

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = 3 + \frac{\cos 4x}{\sin 4x}\cdot 4 + \log 2.

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = 3 + 4\cot 4x + \log 2.

\displaystyle \Rightarrow \frac{dy}{dx} = y[3 + 4\cot 4x + \log 2].

\displaystyle \Rightarrow \frac{dy}{dx} = e^{3x}\sin 4x\,2^x[3 + 4\cot 4x + \log 2] \quad [\text{From (i)}].

\displaystyle \textbf{Question 24: }~y=\sin x\sin 2x\sin 3x\sin 4x
\displaystyle \text{Answer:}

\displaystyle \text{We have } y = \sin x \, \sin 2x \, \sin 3x \, \sin 4x \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides, } \log y = \log(\sin x \sin 2x \sin 3x \sin 4x).

\displaystyle \Rightarrow \log y = \log(\sin x) + \log(\sin 2x) + \log(\sin 3x) + \log(\sin 4x).

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule,}

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}[\log(\sin x)] + \frac{d}{dx}[\log(\sin 2x)] + \frac{d}{dx}[\log(\sin 3x)] + \frac{d}{dx}[\log(\sin 4x)].

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{1}{\sin x}\frac{d}{dx}(\sin x) + \frac{1}{\sin 2x}\frac{d}{dx}(\sin 2x) + \frac{1}{\sin 3x}\frac{d}{dx}(\sin 3x) + \frac{1}{\sin 4x}\frac{d}{dx}(\sin 4x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \cot x + 2\cot 2x + 3\cot 3x + 4\cot 4x.

\displaystyle \Rightarrow \frac{dy}{dx} = y[\cot x + 2\cot 2x + 3\cot 3x + 4\cot 4x].

\displaystyle \Rightarrow \frac{dy}{dx} = \sin x \sin 2x \sin 3x \sin 4x [\cot x + 2\cot 2x + 3\cot 3x + 4\cot 4x] \quad [\text{From (i)}].

\displaystyle \textbf{Question 25: }~y=x^{\sin x}+(\sin x)^x
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = x^{\sin x} + (\sin x)^x.

\displaystyle \text{Also, let } u = x^{\sin x} \text{ and } v = (\sin x)^x.

\displaystyle \therefore y = u + v.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \quad \text{...(i)}.

\displaystyle \text{Now, } u = x^{\sin x}.

\displaystyle \text{Taking log on both sides, } \log u = \log(x^{\sin x}).

\displaystyle \Rightarrow \log u = \sin x \log x.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = \log x \frac{d}{dx}(\sin x) + \sin x \frac{d}{dx}(\log x).

\displaystyle \Rightarrow \frac{du}{dx} = u\left[\cos x \log x + \frac{\sin x}{x}\right].

\displaystyle \Rightarrow \frac{du}{dx} = x^{\sin x}\left[\cos x \log x + \frac{\sin x}{x}\right] \quad \text{...(ii)}.

\displaystyle \text{Again, } v = (\sin x)^x.

\displaystyle \text{Taking log on both sides, } \log v = \log[(\sin x)^x].

\displaystyle \Rightarrow \log v = x \log(\sin x).

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} = \log(\sin x)\frac{d}{dx}(x) + x\frac{d}{dx}[\log(\sin x)].

\displaystyle \Rightarrow \frac{dv}{dx} = v\left[\log(\sin x) + x\cot x\right].

\displaystyle \Rightarrow \frac{dv}{dx} = (\sin x)^x\left[\log(\sin x) + x\cot x\right] \quad \text{...(iii)}.

\displaystyle \text{From (i), (ii) and (iii), we obtain}

\displaystyle \frac{dy}{dx} = x^{\sin x}\left[\cos x \log x + \frac{\sin x}{x}\right] + (\sin x)^x\left[\log(\sin x) + x\cot x\right].

\displaystyle \textbf{Question 26: }~y=(\sin x)^{\cos x}+(\cos x)^{\sin x}
\displaystyle \text{Answer:}

\displaystyle \text{We have } y = (\sin x)^{\cos x} + (\cos x)^{\sin x}.

\displaystyle \Rightarrow y = e^{\cos x \log(\sin x)} + e^{\sin x \log(\cos x)}.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{d}{dx}\!\left(e^{\cos x \log(\sin x)}\right) + \frac{d}{dx}\!\left(e^{\sin x \log(\cos x)}\right).

\displaystyle \Rightarrow \frac{dy}{dx} = e^{\cos x \log(\sin x)}\frac{d}{dx}[\cos x \log(\sin x)] + e^{\sin x \log(\cos x)}\frac{d}{dx}[\sin x \log(\cos x)].

\displaystyle \Rightarrow \frac{dy}{dx} = e^{\cos x \log(\sin x)} \left[\cos x \frac{d}{dx}(\log(\sin x)) + \log(\sin x)\frac{d}{dx}(\cos x)\right]

\displaystyle \qquad + e^{\sin x \log(\cos x)} \left[\sin x \frac{d}{dx}(\log(\cos x)) + \log(\cos x)\frac{d}{dx}(\sin x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (\sin x)^{\cos x} \left[\cos x \cot x - \sin x \log(\sin x)\right] + (\cos x)^{\sin x} \left[\cos x \log(\cos x) - \sin x \tan x\right].

\displaystyle \textbf{Question 27: }~y=(\tan x)^{\cot x}+(\cot x)^{\tan x}
\displaystyle \text{Answer:}

\displaystyle \text{We have } y = (\tan x)^{\cot x} + (\cot x)^{\tan x}.

\displaystyle \Rightarrow y = e^{\cot x \log(\tan x)} + e^{\tan x \log(\cot x)}.

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule,}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{d}{dx}\!\left(e^{\cot x \log(\tan x)}\right) + \frac{d}{dx}\!\left(e^{\tan x \log(\cot x)}\right).

\displaystyle \Rightarrow \frac{dy}{dx} = e^{\cot x \log(\tan x)}\frac{d}{dx}[\cot x \log(\tan x)] + e^{\tan x \log(\cot x)}\frac{d}{dx}[\tan x \log(\cot x)].

\displaystyle \Rightarrow \frac{dy}{dx} = e^{\cot x \log(\tan x)} \left[\cot x\frac{d}{dx}(\log(\tan x)) + \log(\tan x)\frac{d}{dx}(\cot x)\right]

\displaystyle \qquad + e^{\tan x \log(\cot x)} \left[\tan x\frac{d}{dx}(\log(\cot x)) + \log(\cot x)\frac{d}{dx}(\tan x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (\tan x)^{\cot x} \left[\cot x\left(\frac{\sec^2 x}{\tan x}\right) - \log(\tan x)\mathrm{cosec}^2 x\right] + (\cot x)^{\tan x} \left[\tan x\left(\frac{-\mathrm{cosec}^2 x}{\cot x}\right) + \log(\cot x)\sec^2 x\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (\tan x)^{\cot x}\mathrm{cosec}^2 x[1-\log(\tan x)] + (\cot x)^{\tan x}\sec^2 x[\log(\cot x)-1].

\displaystyle \textbf{Question 28: }~y=(\sin x)^x+\sin^{-1}\sqrt{x}
\displaystyle \text{Answer:}

\displaystyle \text{We have } y = (\sin x)^x + \sin^{-1}\!\sqrt{x}.

\displaystyle \text{Let } u = (\sin x)^x \text{ and } v = \sin^{-1}\!\sqrt{x}.

\displaystyle \therefore y = u + v \;\Rightarrow\; \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}.

\displaystyle \text{Now, } u = (\sin x)^x.

\displaystyle \Rightarrow \log u = x\log(\sin x).

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = \log(\sin x) + x\frac{1}{\sin x}\frac{d}{dx}(\sin x).

\displaystyle \Rightarrow \frac{du}{dx} = (\sin x)^x[\log(\sin x) + x\cot x].

\displaystyle \text{Again, } v = \sin^{-1}\!\sqrt{x}.

\displaystyle \Rightarrow \frac{dv}{dx} = \frac{1}{\sqrt{1-(\sqrt{x})^2}}\cdot\frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x(1-x)}}.

\displaystyle \Rightarrow \frac{dy}{dx} = (\sin x)^x[\log(\sin x) + x\cot x] + \frac{1}{2\sqrt{x(1-x)}}.

\displaystyle \textbf{Question 29 (i): }~y=x^{\cos x}+(\sin x)^{\tan x}\;[\text{CBSE 2009}]

\displaystyle \textbf{(i): }~y=x^{\cos x}+(\sin x)^{\tan x}\;[\text{CBSE 2009}]
\displaystyle \text{Answer:}

\displaystyle \text{We have } y = x^{\cos x} + (\sin x)^{\tan x}.

\displaystyle \Rightarrow y = e^{\cos x \log x} + e^{\tan x \log(\sin x)}.

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule,}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{d}{dx}\!\left(e^{\cos x \log x}\right) + \frac{d}{dx}\!\left(e^{\tan x \log(\sin x)}\right).

\displaystyle \Rightarrow \frac{dy}{dx} = e^{\cos x \log x}\frac{d}{dx}[\cos x \log x] + e^{\tan x \log(\sin x)}\frac{d}{dx}[\tan x \log(\sin x)].

\displaystyle \Rightarrow \frac{dy}{dx} = e^{\cos x \log x} \left[\cos x\frac{d}{dx}(\log x) + \log x\frac{d}{dx}(\cos x)\right]

\displaystyle \qquad + e^{\tan x \log(\sin x)} \left[\tan x\frac{d}{dx}(\log(\sin x)) + \log(\sin x)\frac{d}{dx}(\tan x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = x^{\cos x}\left[\frac{\cos x}{x} - \sin x \log x\right] + (\sin x)^{\tan x}\left[\tan x \cot x + \sec^2 x \log(\sin x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = x^{\cos x}\left[\frac{\cos x}{x} - \sin x \log x\right] + (\sin x)^{\tan x}\left[1 + \sec^2 x \log(\sin x)\right].

\displaystyle \textbf{(ii): }~y=x^+(\sin x)^x \;[\text{CBSE 2009}]
\displaystyle \text{Answer:}

\displaystyle \text{We have } y = x^x + (\sin x)^x.

\displaystyle \Rightarrow y = e^{x\log x} + e^{x\log(\sin x)}.

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule and product rule,}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{d}{dx}\!\left(e^{x\log x}\right) + \frac{d}{dx}\!\left(e^{x\log(\sin x)}\right).

\displaystyle \Rightarrow \frac{dy}{dx} = e^{x\log x}\frac{d}{dx}(x\log x) + e^{x\log(\sin x)}\frac{d}{dx}[x\log(\sin x)].

\displaystyle \Rightarrow \frac{dy}{dx} = e^{x\log x}\left[x\frac{d}{dx}(\log x)+\log x\frac{d}{dx}(x)\right] + e^{x\log(\sin x)}\left[x\frac{d}{dx}(\log(\sin x))+\log(\sin x)\frac{d}{dx}(x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = x^x[1+\log x] + (\sin x)^x[x\cot x+\log(\sin x)].

\displaystyle \textbf{Question 30: }~y=(\tan x)^{\log x}+\cos^2\!\left(\frac{\pi}{4}\right)
\displaystyle \text{Answer:}

\displaystyle \text{We have } y = (\tan x)^{\log x} + \cos^2\!\left(\frac{\pi}{4}\right).

\displaystyle \Rightarrow y = e^{\log x \log(\tan x)} + \cos^2\!\left(\frac{\pi}{4}\right).

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule,}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{d}{dx}\!\left(e^{\log x \log(\tan x)}\right) + \frac{d}{dx}\!\left[\cos^2\!\left(\frac{\pi}{4}\right)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = e^{\log x \log(\tan x)}\frac{d}{dx}[\log x \log(\tan x)] + 0.

\displaystyle \Rightarrow \frac{dy}{dx} = e^{\log x \log(\tan x)} \left[\log x\frac{d}{dx}(\log(\tan x)) + \log(\tan x)\frac{d}{dx}(\log x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = (\tan x)^{\log x} \left[\log x\left(\frac{\sec^2 x}{\tan x}\right) + \frac{\log(\tan x)}{x}\right].

\displaystyle \textbf{Question 31: }~y=x^x+x^{1/x}
\displaystyle \text{Answer:}

\displaystyle \text{We have } y = x^x + x^{1/x}.

\displaystyle \Rightarrow y = e^{x\log x} + e^{\frac{1}{x}\log x}.

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule,}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{d}{dx}\!\left(e^{x\log x}\right) + \frac{d}{dx}\!\left(e^{\frac{1}{x}\log x}\right).

\displaystyle \Rightarrow \frac{dy}{dx} = e^{x\log x}\frac{d}{dx}(x\log x) + e^{\frac{1}{x}\log x}\frac{d}{dx}\!\left(\frac{1}{x}\log x\right).

\displaystyle \Rightarrow \frac{dy}{dx} = e^{x\log x}\left[x\frac{d}{dx}(\log x)+\log x\frac{d}{dx}(x)\right] + e^{\frac{1}{x}\log x}\left[\frac{1}{x}\frac{d}{dx}(\log x)+\log x\frac{d}{dx}\!\left(\frac{1}{x}\right)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = x^x(1+\log x) + x^{1/x}\left(\frac{1}{x^2}-\frac{\log x}{x^2}\right).

\displaystyle \Rightarrow \frac{dy}{dx} = x^x(1+\log x) + x^{1/x}\left(\frac{1-\log x}{x^2}\right).

\displaystyle \textbf{Question 32: }~y=x^{\log x}+(\log x)^x
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = x^{\log x} + (\log x)^x.

\displaystyle \text{Also, let } u = (\log x)^x \text{ and } v = x^{\log x}.

\displaystyle \therefore y = u + v.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \quad \text{...(i)}.

\displaystyle \text{Now, } u = (\log x)^x.

\displaystyle \Rightarrow \log u = \log[(\log x)^x].

\displaystyle \Rightarrow \log u = x\log(\log x).

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = \log(\log x)\frac{d}{dx}(x) + x\frac{d}{dx}[\log(\log x)].

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = \log(\log x) + \frac{x}{\log x}\frac{d}{dx}(\log x).

\displaystyle \Rightarrow \frac{du}{dx} = (\log x)^x\left[\log(\log x) + \frac{1}{\log x}\right] \quad \text{...(ii)}.

\displaystyle \text{Again, } v = x^{\log x}.

\displaystyle \Rightarrow \log v = \log(x^{\log x}).

\displaystyle \Rightarrow \log v = (\log x)(\log x) = (\log x)^2.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} = 2(\log x)\frac{d}{dx}(\log x).

\displaystyle \Rightarrow \frac{dv}{dx} = 2x^{\log x}\frac{\log x}{x} \quad \text{...(iii)}.

\displaystyle \text{From (i), (ii) and (iii), we obtain}

\displaystyle \frac{dy}{dx} = 2x^{\log x}\frac{\log x}{x} + (\log x)^x\left[\log(\log x) + \frac{1}{\log x}\right].

\displaystyle \textbf{Question 33: }~\text{If }x^{13}y^7=(x+y)^{20},\text{ prove that }\frac{dy}{dx}=\frac{y}{x}
\displaystyle \text{Answer:}

\displaystyle \text{We have } x^{13}y^{7} = (x+y)^{20}.

\displaystyle \text{Taking log on both sides,}

\displaystyle \log(x^{13}y^{7}) = \log[(x+y)^{20}].

\displaystyle \Rightarrow 13\log x + 7\log y = 20\log(x+y).

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule,}

\displaystyle \Rightarrow 13\frac{1}{x} + 7\frac{1}{y}\frac{dy}{dx} = 20\frac{1}{x+y}\frac{d}{dx}(x+y).

\displaystyle \Rightarrow 13\frac{1}{x} + 7\frac{1}{y}\frac{dy}{dx} = \frac{20}{x+y}\left(1+\frac{dy}{dx}\right).

\displaystyle \Rightarrow \frac{7}{y}\frac{dy}{dx} - \frac{20}{x+y}\frac{dy}{dx} = \frac{20}{x+y} - \frac{13}{x}.

\displaystyle \Rightarrow \frac{dy}{dx} \left[\frac{7}{y} - \frac{20}{x+y}\right] = \frac{20}{x+y} - \frac{13}{x}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{20x - 13(x+y)}{x(x+y)} \cdot \frac{y(x+y)}{7(x+y) - 20y}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y(7x - 13y)}{x(7x - 13y)}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y}{x}.

\displaystyle \textbf{Question 34: }~\text{If }x^{16}y^9=(x^2+y)^{17},\text{ prove that }x\frac{dy}{dx}=2y
\displaystyle \text{Answer:}

\displaystyle \text{We have } x^{16}y^{9} = (x^2+y)^{17}.

\displaystyle \text{Taking log on both sides,}

\displaystyle \log(x^{16}y^{9}) = \log[(x^2+y)^{17}].

\displaystyle \Rightarrow 16\log x + 9\log y = 17\log(x^2+y).

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule,}

\displaystyle \Rightarrow 16\frac{1}{x} + 9\frac{1}{y}\frac{dy}{dx} = 17\frac{1}{x^2+y}\frac{d}{dx}(x^2+y).

\displaystyle \Rightarrow 16\frac{1}{x} + 9\frac{1}{y}\frac{dy}{dx} = \frac{17}{x^2+y}\left(2x+\frac{dy}{dx}\right).

\displaystyle \Rightarrow \frac{9}{y}\frac{dy}{dx} - \frac{17}{x^2+y}\frac{dy}{dx} = \frac{34x}{x^2+y} - \frac{16}{x}.

\displaystyle \Rightarrow \frac{dy}{dx} \left[\frac{9}{y} - \frac{17}{x^2+y}\right] = \frac{34x}{x^2+y} - \frac{16}{x}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{34x(x) - 16(x^2+y)}{x(x^2+y)} \cdot \frac{y(x^2+y)}{9(x^2+y)-17y}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y(18x^2-16y)}{x(9x^2-8y)}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{2y}{x}.

\displaystyle \Rightarrow x\frac{dy}{dx} = 2y.

\displaystyle \textbf{Question 35: }~\text{If }y=\sin\left(x^x\right),\text{ prove that }\frac{dy}{dx}=\cos\left(x^x\right)\cdot x^x(1+\log x)
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = \sin(x^x) \quad \text{...(i)}.

\displaystyle \text{Also, let } u = x^x \quad \text{...(ii)}.

\displaystyle \text{Taking log on both sides,}

\displaystyle \Rightarrow \log u = \log(x^x).

\displaystyle \Rightarrow \log u = x\log x.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = \frac{d}{dx}(x\log x).

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = x\frac{d}{dx}(\log x) + \log x\frac{d}{dx}(x).

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = x\left(\frac{1}{x}\right) + \log x.

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = 1 + \log x.

\displaystyle \Rightarrow \frac{du}{dx} = u(1+\log x).

\displaystyle \Rightarrow \frac{du}{dx} = x^x(1+\log x) \quad \text{...(iii)}.

\displaystyle \text{Now, using (ii) in (i), } y = \sin u.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow \frac{dy}{dx} = \cos u \frac{du}{dx}.

\displaystyle \Rightarrow \frac{dy}{dx} = \cos(x^x)\,x^x(1+\log x).

\displaystyle \textbf{Question 36: }~\text{If }x^x+y^x=1,\text{ prove that }\frac{dy}{dx}=-\left[\frac{x^x(1+\log x)+y^x\log y}{x\cdot y^{(x-1)}}\right]
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x^x + y^x = 1.

\displaystyle \Rightarrow e^{\log x^x} + e^{\log y^x} = 1.

\displaystyle \Rightarrow e^{x\log x} + e^{x\log y} = 1.

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule,}

\displaystyle \frac{d}{dx}\left(e^{x\log x}\right) + \frac{d}{dx}\left(e^{x\log y}\right) = \frac{d}{dx}(1).

\displaystyle e^{x\log x}\frac{d}{dx}(x\log x) + e^{x\log y}\frac{d}{dx}(x\log y) = 0.

\displaystyle e^{x\log x} \left[x\frac{d}{dx}(\log x)+\log x\frac{d}{dx}(x)\right] + e^{x\log y} \left[x\frac{d}{dx}(\log y)+\log y\frac{d}{dx}(x)\right] = 0.

\displaystyle x^x\left[x\left(\frac{1}{x}\right)+\log x\right] + y^x\left[x\left(\frac{1}{y}\frac{dy}{dx}\right)+\log y\right] = 0.

\displaystyle x^x(1+\log x) + y^x\left(\frac{x}{y}\frac{dy}{dx}+\log y\right) = 0.

\displaystyle y^x\left(\frac{x}{y}\frac{dy}{dx}\right) = -\left[x^x(1+\log x)+y^x\log y\right].

\displaystyle \frac{x}{y}y^x\frac{dy}{dx} = -\left[x^x(1+\log x)+y^x\log y\right].

\displaystyle \frac{dy}{dx} = -\frac{x^x(1+\log x)+y^x\log y}{x\,y^{\,x-1}}.

\displaystyle \textbf{Question 37: }~\text{If }x^y\cdot y^x=1,\text{ prove that }\frac{dy}{dx}=-\frac{y(y+x\log y)}{x(y\log x+x)}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x^y \times y^x = 1.

\displaystyle \text{Taking log on both sides,}

\displaystyle \log(x^y \times y^x) = \log 1.

\displaystyle \Rightarrow y\log x + x\log y = 0.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \frac{d}{dx}(y\log x) + \frac{d}{dx}(x\log y) = 0.

\displaystyle \left[y\frac{d}{dx}(\log x) + \log x\frac{dy}{dx}\right] + \left[x\frac{d}{dx}(\log y) + \log y\frac{d}{dx}(x)\right] = 0.

\displaystyle \left[y\left(\frac{1}{x}\right) + \log x\frac{dy}{dx}\right] + \left[x\left(\frac{1}{y}\frac{dy}{dx}\right) + \log y\right] = 0.

\displaystyle \frac{y}{x} + \log x\frac{dy}{dx} + \frac{x}{y}\frac{dy}{dx} + \log y = 0.

\displaystyle \frac{dy}{dx}\left(\log x + \frac{x}{y}\right) = -\left(\log y + \frac{y}{x}\right).

\displaystyle \frac{dy}{dx} = -\frac{\log y + \frac{y}{x}}{\log x + \frac{x}{y}}.

\displaystyle \frac{dy}{dx} = -\frac{y(x\log y + y)}{x(y\log x + x)}.

\displaystyle \textbf{Question 38: }~\text{If }x^y+y^x=(x+y)^{x+y},\text{ find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x^y + y^x = (x+y)^{x+y}.

\displaystyle \Rightarrow e^{y\log x} + e^{x\log y} = e^{(x+y)\log(x+y)}.

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule and product rule,}

\displaystyle \frac{d}{dx}\!\left(e^{y\log x}\right) + \frac{d}{dx}\!\left(e^{x\log y}\right) = \frac{d}{dx}\!\left(e^{(x+y)\log(x+y)}\right).

\displaystyle e^{y\log x}\!\left[ y\frac{d}{dx}(\log x) + \log x\frac{dy}{dx} \right] + e^{x\log y}\!\left[ x\frac{d}{dx}(\log y) + \log y\frac{d}{dx}(x) \right]

\displaystyle = e^{(x+y)\log(x+y)} \frac{d}{dx}\!\left[(x+y)\log(x+y)\right].

\displaystyle x^y\!\left[\frac{y}{x} + \log x\frac{dy}{dx}\right] + y^x\!\left[\frac{x}{y}\frac{dy}{dx} + \log y\right]

\displaystyle = (x+y)^{x+y} \left[(1+\frac{dy}{dx}) + \log(x+y)(1+\frac{dy}{dx})\right].

\displaystyle x^y\frac{y}{x} + x^y\log x\frac{dy}{dx} + y^x\frac{x}{y}\frac{dy}{dx} + y^x\log y

\displaystyle = (x+y)^{x+y}\!\left[1+\log(x+y)\right] + (x+y)^{x+y}\!\left[1+\log(x+y)\right]\frac{dy}{dx}.

\displaystyle \frac{dy}{dx} \left[x^y\log x + y^x\frac{x}{y} - (x+y)^{x+y}\!\left(1+\log(x+y)\right)\right]

\displaystyle = (x+y)^{x+y}\!\left(1+\log(x+y)\right) - x^{y-1}y - y^x\log y.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{(x+y)^{x+y}\!\left[1+\log(x+y)\right] - x^{y-1}y - y^x\log y} {x^y\log x + x y^{x-1} - (x+y)^{x+y}\!\left[1+\log(x+y)\right]}.

\displaystyle \textbf{Question 39: }~\text{If }x^m y^n=1,\text{ prove that }\frac{dy}{dx}=-\frac{my}{nx}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x^{m}y^{n} = 1.

\displaystyle \text{Taking log on both sides,}

\displaystyle \log(x^{m}y^{n}) = \log 1.

\displaystyle \Rightarrow m\log x + n\log y = 0.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \Rightarrow m\frac{d}{dx}(\log x) + n\frac{d}{dx}(\log y) = 0.

\displaystyle \Rightarrow m\left(\frac{1}{x}\right) + n\left(\frac{1}{y}\frac{dy}{dx}\right) = 0.

\displaystyle \Rightarrow \frac{n}{y}\frac{dy}{dx} = -\frac{m}{x}.

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{m}{n}\frac{y}{x}.

\displaystyle \textbf{Question 40: }~\text{If }y^x=e^{\,y-x},\text{ prove that }\frac{dy}{dx}=\frac{(1+\log y)^2}{\log y}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } y^{x} = e^{\,y-x}.

\displaystyle \text{Taking log on both sides,}

\displaystyle \log(y^{x}) = \log\!\left(e^{\,y-x}\right).

\displaystyle \Rightarrow x\log y = (y-x)\log e.

\displaystyle \Rightarrow x\log y = y - x \quad \text{...(i)}.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \frac{d}{dx}(x\log y) = \frac{d}{dx}(y-x).

\displaystyle x\frac{d}{dx}(\log y) + \log y\frac{d}{dx}(x) = \frac{dy}{dx} - 1.

\displaystyle x\left(\frac{1}{y}\frac{dy}{dx}\right) + \log y = \frac{dy}{dx} - 1.

\displaystyle \frac{dy}{dx}\left(\frac{x}{y} - 1\right) = -\,(1+\log y).

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{1+\log y}{\frac{x}{y}-1}.

\displaystyle \text{Using (i): } x = \frac{y}{1+\log y}.

\displaystyle \Rightarrow \frac{x}{y}-1 = \frac{1}{1+\log y}-1 = -\frac{\log y}{1+\log y}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{(1+\log y)^2}{\log y}.

\displaystyle \textbf{Question 41: }~\text{If }(\sin x)^y=(\cos y)^x,\text{ prove that }\frac{dy}{dx}=\frac{\log(\cos y)-y\cot x}{\log(\sin x)+x\tan y}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } (\sin x)^y = (\cos y)^x.

\displaystyle \text{Taking log on both sides,}

\displaystyle \log(\sin x)^y = \log(\cos y)^x.

\displaystyle \Rightarrow y\log(\sin x) = x\log(\cos y).

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \frac{d}{dx}\!\left[y\log(\sin x)\right] = \frac{d}{dx}\!\left[x\log(\cos y)\right].

\displaystyle y\frac{d}{dx}(\log(\sin x)) + \log(\sin x)\frac{dy}{dx} = x\frac{d}{dx}(\log(\cos y)) + \log(\cos y)\frac{d}{dx}(x).

\displaystyle y\left(\frac{1}{\sin x}\cos x\right) + \log(\sin x)\frac{dy}{dx} = x\left(\frac{1}{\cos y}(-\sin y)\frac{dy}{dx}\right) + \log(\cos y).

\displaystyle y\cot x + \log(\sin x)\frac{dy}{dx} = -x\tan y\,\frac{dy}{dx} + \log(\cos y).

\displaystyle \frac{dy}{dx}\left[\log(\sin x) + x\tan y\right] = \log(\cos y) - y\cot x.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{\log(\cos y) - y\cot x}{\log(\sin x) + x\tan y}.

\displaystyle \textbf{Question 42: }~\text{If }(\cos x)^y=(\tan y)^x,\text{ prove that }\frac{dy}{dx}=\frac{\log(\tan y)+y\tan x}{\log(\cos x)-x\sec y\mathrm{cosec} y}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } (\cos x)^y = (\tan y)^x.

\displaystyle \text{Taking log on both sides,}

\displaystyle \log(\cos x)^y = \log(\tan y)^x.

\displaystyle \Rightarrow y\log(\cos x) = x\log(\tan y).

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule,}

\displaystyle \frac{d}{dx}\!\left[y\log(\cos x)\right] = \frac{d}{dx}\!\left[x\log(\tan y)\right].

\displaystyle y\frac{d}{dx}(\log(\cos x)) + \log(\cos x)\frac{dy}{dx} = x\frac{d}{dx}(\log(\tan y)) + \log(\tan y)\frac{d}{dx}(x).

\displaystyle y\left(\frac{-\sin x}{\cos x}\right) + \log(\cos x)\frac{dy}{dx} = x\left(\frac{1}{\tan y}\sec^2 y\,\frac{dy}{dx}\right) + \log(\tan y).

\displaystyle -\,y\tan x + \log(\cos x)\frac{dy}{dx} = x\sec y\mathrm{cosec} y\,\frac{dy}{dx} + \log(\tan y).

\displaystyle \frac{dy}{dx}\!\left[\log(\cos x) - x\sec y\mathrm{cosec} y\right] = \log(\tan y) + y\tan x.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{\log(\tan y) + y\tan x} {\log(\cos x) - x\sec y\mathrm{cosec} y}.

\displaystyle \textbf{Question 43: }~\text{If }e^x+e^y=e^{x+y},\text{ prove that }\frac{dy}{dx}+e^{\,y-x}=0
\displaystyle \text{Answer:}

\displaystyle \text{We have, } e^{x} + e^{y} = e^{x+y}.

\displaystyle \text{Differentiating both sides with respect to } x \text{ using chain rule,}

\displaystyle \frac{d}{dx}(e^{x}) + \frac{d}{dx}(e^{y}) = \frac{d}{dx}(e^{x+y}).

\displaystyle e^{x} + e^{y}\frac{dy}{dx} = e^{x+y}\frac{d}{dx}(x+y).

\displaystyle e^{x} + e^{y}\frac{dy}{dx} = e^{x+y}\left(1+\frac{dy}{dx}\right).

\displaystyle e^{y}\frac{dy}{dx} - e^{x+y}\frac{dy}{dx} = e^{x+y} - e^{x}.

\displaystyle \frac{dy}{dx}(e^{y} - e^{x+y}) = e^{x+y} - e^{x}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{e^{x+y} - e^{x}}{e^{y} - e^{x+y}}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{e^{x}(e^{y}-1)}{e^{y}(1-e^{x})}. \text{      Using equation 1}

\displaystyle \Rightarrow \frac{dy}{dx} = -\,\frac{e^{x}}{e^{y}}.

\displaystyle \textbf{Question 44: }~\text{If }e^y=y^x,\text{ prove that }\frac{dy}{dx}=\frac{(\log y)^2}{\log y-1}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } e^{y} = y^{x}.

\displaystyle \text{Taking log on both sides,}

\displaystyle \log(e^{y}) = \log(y^{x}).

\displaystyle \Rightarrow y\log e = x\log y.

\displaystyle \Rightarrow y = x\log y \quad \text{...(i)}.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \frac{dy}{dx} = \frac{d}{dx}(x\log y).

\displaystyle \Rightarrow \frac{dy}{dx} = x\frac{d}{dx}(\log y) + \log y\frac{d}{dx}(x).

\displaystyle \Rightarrow \frac{dy}{dx} = x\left(\frac{1}{y}\frac{dy}{dx}\right) + \log y.

\displaystyle \Rightarrow \frac{dy}{dx}\left(1 - \frac{x}{y}\right) = \log y.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y\log y}{\,y-x\,}.

\displaystyle \text{Using (i): } x = \frac{y}{\log y}.

\displaystyle \Rightarrow y-x = y\left(1-\frac{1}{\log y}\right) = \frac{y(\log y - 1)}{\log y}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y\log y}{\frac{y(\log y - 1)}{\log y}}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{(\log y)^2}{\log y - 1}.

\displaystyle \textbf{Question 45: }~\text{If }e^{x+y}-x=0,\text{ prove that }\frac{dy}{dx}=\frac{1-x}{x}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } e^{x+y} - x = 0.

\displaystyle \Rightarrow e^{x+y} = x \quad \text{...(i)}.

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule,}

\displaystyle \frac{d}{dx}\!\left(e^{x+y}\right) = \frac{d}{dx}(x).

\displaystyle e^{x+y}\frac{d}{dx}(x+y) = 1.

\displaystyle e^{x+y}\left(1+\frac{dy}{dx}\right) = 1.

\displaystyle \text{Using (i), } x\left(1+\frac{dy}{dx}\right) = 1.

\displaystyle \Rightarrow 1+\frac{dy}{dx} = \frac{1}{x}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{x} - 1.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1-x}{x}.

\displaystyle \textbf{Question 46: }~\text{If }y=x\sin(a+y),\text{ prove that }\frac{dy}{dx}=\frac{\sin^2(a+y)}{\sin(a+y)-y\cos(a+y)}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } y = x\sin(a+y).

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule,}

\displaystyle \frac{dy}{dx} = x\frac{d}{dx}\bigl[\sin(a+y)\bigr] + \sin(a+y)\frac{d}{dx}(x).

\displaystyle \Rightarrow \frac{dy}{dx} = x\cos(a+y)\frac{dy}{dx} + \sin(a+y).

\displaystyle \Rightarrow \left[1 - x\cos(a+y)\right]\frac{dy}{dx} = \sin(a+y).

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{\sin(a+y)}{1 - x\cos(a+y)}.

\displaystyle \text{Using } y = x\sin(a+y) \Rightarrow x = \frac{y}{\sin(a+y)}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{\sin(a+y)}{1 - \frac{y}{\sin(a+y)}\cos(a+y)}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{\sin^2(a+y)}{\sin(a+y) - y\cos(a+y)}.

\displaystyle \textbf{Question 47: }~\text{If }x\sin(a+y)+\sin a\cos(a+y)=0,\text{ prove that }\frac{dy}{dx}=\frac{\sin^2(a+y)}{\sin a}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x\sin(a+y) + \sin a \cos(a+y) = 0.

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule,}

\displaystyle \frac{d}{dx}\!\left[x\sin(a+y) + \sin a \cos(a+y)\right] = 0.

\displaystyle x\frac{d}{dx}[\sin(a+y)] + \sin(a+y)\frac{d}{dx}(x) + \sin a\frac{d}{dx}[\cos(a+y)] = 0.

\displaystyle x\cos(a+y)\frac{dy}{dx} + \sin(a+y) - \sin a\sin(a+y)\frac{dy}{dx} = 0.

\displaystyle \left[x\cos(a+y) - \sin a\sin(a+y)\right]\frac{dy}{dx} + \sin(a+y) = 0.

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{\sin(a+y)}{x\cos(a+y) - \sin a\sin(a+y)}.

\displaystyle \text{From the given equation, } x = -\frac{\sin a\cos(a+y)}{\sin(a+y)}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{\sin(a+y)}{\sin a\!\left[\cos^2(a+y)+\sin^2(a+y)\right]}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{\sin^2(a+y)}{\sin a}.

\displaystyle \textbf{Question 48: }~\text{If }(\sin x)^y=x+y,\text{ prove that }\frac{dy}{dx}=\frac{1-(x+y)y\cot x}{(x+y)\log(\sin x)-1}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } (\sin x)^y = x + y.

\displaystyle \text{Taking log on both sides,}

\displaystyle \log(\sin x)^y = \log(x+y).

\displaystyle \Rightarrow y\log(\sin x) = \log(x+y).

\displaystyle \text{Differentiating with respect to } x \text{ using chain rule,}

\displaystyle \frac{d}{dx}\!\left[y\log(\sin x)\right] = \frac{d}{dx}\!\left[\log(x+y)\right].

\displaystyle y\frac{d}{dx}(\log(\sin x)) + \log(\sin x)\frac{dy}{dx} = \frac{1}{x+y}\frac{d}{dx}(x+y).

\displaystyle y\left(\frac{1}{\sin x}\cos x\right) + \log(\sin x)\frac{dy}{dx} = \frac{1}{x+y}\left(1+\frac{dy}{dx}\right).

\displaystyle y\cot x + \log(\sin x)\frac{dy}{dx} = \frac{1}{x+y} + \frac{1}{x+y}\frac{dy}{dx}.

\displaystyle \frac{dy}{dx}\left[\log(\sin x) - \frac{1}{x+y}\right] = \frac{1}{x+y} - y\cot x.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1 - y(x+y)\cot x}{(x+y)\log(\sin x) - 1}.

\displaystyle \textbf{Question 49: }~\text{If }xy\log(x+y)=1,\text{ prove that }\frac{dy}{dx}=-\frac{y(x^2y+x+y)}{x(xy^2+x+y)}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } xy\log(x+y)=1 \quad \text{...(i)}.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \frac{d}{dx}\!\left[xy\log(x+y)\right]=\frac{d}{dx}(1).

\displaystyle xy\frac{d}{dx}\!\left[\log(x+y)\right] + \log(x+y)\frac{d}{dx}(xy)=0.

\displaystyle xy\left(\frac{1}{x+y}\right)\!\left(1+\frac{dy}{dx}\right) + \log(x+y)\left(x\frac{dy}{dx}+y\right)=0.

\displaystyle \frac{xy}{x+y} + \frac{xy}{x+y}\frac{dy}{dx} + x\log(x+y)\frac{dy}{dx} + y\log(x+y)=0.

\displaystyle \frac{dy}{dx}\!\left[\frac{xy}{x+y}+x\log(x+y)\right] = -\left[\frac{xy}{x+y}+y\log(x+y)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{\frac{xy}{x+y}+y\log(x+y)} {\frac{xy}{x+y}+x\log(x+y)}.

\displaystyle \text{Using (i): } \log(x+y)=\frac{1}{xy}.

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{\frac{xy}{x+y}+\frac{y}{xy}} {\frac{xy}{x+y}+\frac{x}{xy}}.

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{y(x+y+x^{2}y)}{x(x+y+xy^{2})}.

\displaystyle \textbf{Question 50: }~\text{If }y=x\sin y,\text{ prove that }\frac{dy}{dx}=\frac{y}{x(1-x\cos y)}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } y = x\sin y \quad \text{...(i)}.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \frac{dy}{dx} = \frac{d}{dx}(x\sin y).

\displaystyle \Rightarrow \frac{dy}{dx} = x\frac{d}{dx}(\sin y) + \sin y\frac{d}{dx}(x).

\displaystyle \Rightarrow \frac{dy}{dx} = x\cos y\frac{dy}{dx} + \sin y.

\displaystyle \Rightarrow \frac{dy}{dx} - x\cos y\frac{dy}{dx} = \sin y.

\displaystyle \Rightarrow \frac{dy}{dx}(1 - x\cos y) = \sin y.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{\sin y}{1 - x\cos y}.

\displaystyle \text{Using (i): } \sin y = \frac{y}{x}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y}{x(1 - x\cos y)}.

\displaystyle \textbf{Question 51: }~\text{Find the derivative of the function }f(x)\text{ given by }f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)\text{ and hence find }f'(1)
\displaystyle \text{Answer:}

\displaystyle \text{We have, } f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8).

\displaystyle \text{Taking log on both sides,}

\displaystyle \log f(x)=\log(1+x)+\log(1+x^2)+\log(1+x^4)+\log(1+x^8).

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \frac{1}{f(x)}f'(x) =\frac{1}{1+x} +\frac{2x}{1+x^2} +\frac{4x^3}{1+x^4} +\frac{8x^7}{1+x^8}.

\displaystyle \Rightarrow f'(x) =(1+x)(1+x^2)(1+x^4)(1+x^8) \left( \frac{1}{1+x} +\frac{2x}{1+x^2} +\frac{4x^3}{1+x^4} +\frac{8x^7}{1+x^8} \right).

\displaystyle \text{At } x=1,

\displaystyle f'(1) =(1+1)(1+1^2)(1+1^4)(1+1^8) \left( \frac{1}{1+1} +\frac{2(1)}{1+1^2} +\frac{4(1)^3}{1+1^4} +\frac{8(1)^7}{1+1^8} \right).

\displaystyle f'(1)=2\cdot2\cdot2\cdot2 \left( \frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2} \right).

\displaystyle f'(1)=16\left(\frac{1}{2}+1+2+4\right).

\displaystyle f'(1)=16\times\frac{15}{2}=120.

\displaystyle \textbf{Question 52: }~\text{If }y=\log\!\left(\frac{x^2+x+1}{x^2-x+1}\right)+\frac{2}{\sqrt{3}}\tan^{-1}\!\left(\frac{\sqrt{3}\,x}{1-x^2}\right),\text{ find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=\log\!\left(\frac{x^{2}+x+1}{x^{2}-x+1}\right) +\frac{2}{\sqrt3}\tan^{-1}\!\left(\frac{\sqrt3\,x}{1-x^{2}}\right).

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\!\left[\log\!\left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)\right] +\frac{2}{\sqrt3}\frac{d}{dx} \left[\tan^{-1}\!\left(\frac{\sqrt3\,x}{1-x^{2}}\right)\right].

\displaystyle \Rightarrow \frac{dy}{dx} =\frac{1}{\frac{x^{2}+x+1}{x^{2}-x+1}} \frac{d}{dx}\!\left(\frac{x^{2}+x+1}{x^{2}-x+1}\right) +\frac{2}{\sqrt3}\cdot \frac{1}{1+\left(\frac{\sqrt3\,x}{1-x^{2}}\right)^{2}} \frac{d}{dx}\!\left(\frac{\sqrt3\,x}{1-x^{2}}\right).

\displaystyle \Rightarrow \frac{dy}{dx} =\frac{(x^{2}-x+1)(2x+1)-(x^{2}+x+1)(2x-1)} {(x^{2}+x+1)(x^{2}-x+1)} +\frac{2}{\sqrt3}\cdot \frac{(1-x^{2})^{2}} {1+x^{2}+x^{4}} \cdot \frac{\sqrt3(1+x^{2})}{(1-x^{2})^{2}}.

\displaystyle \Rightarrow \frac{dy}{dx} =\frac{2(1-x^{2})}{1+x^{2}+x^{4}} +\frac{2(x^{2}+1)}{1+x^{2}+x^{4}}.

\displaystyle \Rightarrow \frac{dy}{dx} =\frac{2(1-x^{2}+x^{2}+1)}{1+x^{2}+x^{4}}.

\displaystyle \Rightarrow \frac{dy}{dx} =\frac{4}{1+x^{2}+x^{4}}.

\displaystyle \textbf{Question 53: }~\text{If }y=(\sin x-\cos x)^{\sin x-\cos x},~\frac{\pi}{4}<x<\frac{3\pi}{4},\text{ find }\frac{dy}{dx}\;[\text{CBSE 2010}]
\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=(\sin x-\cos x)^{(\sin x-\cos x)} \quad \text{...(i)}.

\displaystyle \text{Taking log on both sides,}

\displaystyle \log y=\log\!\left[(\sin x-\cos x)^{(\sin x-\cos x)}\right].

\displaystyle \Rightarrow \log y=(\sin x-\cos x)\log(\sin x-\cos x).

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}\!\left[(\sin x-\cos x)\log(\sin x-\cos x)\right].

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = (\cos x+\sin x)\log(\sin x-\cos x) +(\sin x-\cos x)\frac{1}{\sin x-\cos x}\frac{d}{dx}(\sin x-\cos x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = (\cos x+\sin x)\log(\sin x-\cos x)+(\cos x+\sin x).

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = (\cos x+\sin x)\left[1+\log(\sin x-\cos x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} = y(\cos x+\sin x)\left[1+\log(\sin x-\cos x)\right].

\displaystyle \Rightarrow \frac{dy}{dx} =(\sin x-\cos x)^{(\sin x-\cos x)} (\cos x+\sin x)\left[1+\log(\sin x-\cos x)\right].

\displaystyle \textbf{Question 54: }~\text{If }xy=e^{x-y},\text{ find }\frac{dy}{dx}\;
\displaystyle \text{Answer:}

\displaystyle \text{We have, } xy = e^{\,x-y}.

\displaystyle \text{Taking log on both sides,}

\displaystyle \log(xy) = \log\!\left(e^{\,x-y}\right).

\displaystyle \Rightarrow \log x + \log y = (x-y)\log e.

\displaystyle \Rightarrow \log x + \log y = x - y.

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \frac{d}{dx}(\log x) + \frac{d}{dx}(\log y) = \frac{d}{dx}(x) - \frac{d}{dx}(y).

\displaystyle \Rightarrow \frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 1 - \frac{dy}{dx}.

\displaystyle \Rightarrow \left(1+\frac{1}{y}\right)\frac{dy}{dx} = 1 - \frac{1}{x}.

\displaystyle \Rightarrow \frac{y+1}{y}\frac{dy}{dx} = \frac{x-1}{x}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y(x-1)}{x(y+1)}.

\displaystyle \textbf{Question 55: }~\text{If }y^x+x^y+x^x=a^b,\text{ find }\frac{dy}{dx}\;
\displaystyle \text{Answer:}

\displaystyle \text{Given that } y^{x} + x^{y} + x^{x} = a^{b}

\displaystyle \text{Putting } u = y^{x},\ v = x^{y} \text{ and } w = x^{x}, \text{ we get}

\displaystyle u + v + w = a^{b}

\displaystyle \therefore \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0 \quad \text{...(i)}

\displaystyle \text{Now, } u = y^{x}

\displaystyle \text{Taking log on both sides,}

\displaystyle \log u = x \log y

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = x \frac{d}{dx}(\log y) + \log y \frac{d}{dx}(x)

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = x \frac{1}{y}\frac{dy}{dx} + \log y

\displaystyle \Rightarrow \frac{du}{dx} = u\left(\frac{x}{y}\frac{dy}{dx} + \log y\right)

\displaystyle \Rightarrow \frac{du}{dx} = y^{x}\left(\frac{x}{y}\frac{dy}{dx} + \log y\right) \quad \text{...(ii)}

\displaystyle \text{Also, } v = x^{y}

\displaystyle \text{Taking log on both sides,}

\displaystyle \log v = y \log x

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} = y \frac{d}{dx}(\log x) + \log x \frac{dy}{dx}

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} = \frac{y}{x} + \log x \frac{dy}{dx}

\displaystyle \Rightarrow \frac{dv}{dx} = v\left(\frac{y}{x} + \log x \frac{dy}{dx}\right)

\displaystyle \Rightarrow \frac{dv}{dx} = x^{y}\left(\frac{y}{x} + \log x \frac{dy}{dx}\right) \quad \text{...(iii)}

\displaystyle \text{Again, } w = x^{x}

\displaystyle \text{Taking log on both sides,}

\displaystyle \log w = x \log x

\displaystyle \Rightarrow \frac{1}{w}\frac{dw}{dx} = x \frac{1}{x} + \log x \cdot 1

\displaystyle \Rightarrow \frac{dw}{dx} = w(1 + \log x)

\displaystyle \Rightarrow \frac{dw}{dx} = x^{x}(1 + \log x) \quad \text{...(iv)}

\displaystyle \text{From (i), (ii), (iii) and (iv), we have}

\displaystyle y^{x}\left(\frac{x}{y}\frac{dy}{dx} + \log y\right) + x^{y}\left(\frac{y}{x} + \log x \frac{dy}{dx}\right) + x^{x}(1 + \log x) = 0

\displaystyle \Rightarrow \left(xy^{x-1} + x^{y}\log x\right)\frac{dy}{dx} = -\left(y^{x}\log y + yx^{y-1} + x^{x}(1 + \log x)\right)

\displaystyle \therefore \frac{dy}{dx} = -\frac{y^{x}\log y + yx^{y-1} + x^{x}(1 + \log x)} {xy^{x-1} + x^{y}\log x}

\displaystyle \textbf{Question 56: }~\text{If }(\cos x)^y=(\cos y)^x,\text{ find }\frac{dy}{dx}\;[\text{CBSE 2012}]
\displaystyle \text{Answer:}

\displaystyle (\cos x)^{y} = (\cos y)^{x}

\displaystyle \text{Taking log on both sides, we get}

\displaystyle y \log(\cos x) = x \log(\cos y)

\displaystyle \Rightarrow \frac{dy}{dx}\log(\cos x) + y \frac{d}{dx}[\log(\cos x)] = \log(\cos y) + x \frac{d}{dx}[\log(\cos y)]

\displaystyle \Rightarrow \frac{dy}{dx}\log(\cos x) - y \tan x = \log(\cos y) - x \tan y \frac{dy}{dx}

\displaystyle \Rightarrow \frac{dy}{dx}\log(\cos x) + x \tan y \frac{dy}{dx} = \log(\cos y) + y \tan x

\displaystyle \Rightarrow \frac{dy}{dx} \left(\log(\cos x) + x \tan y\right) = \log(\cos y) + y \tan x

\displaystyle \therefore \frac{dy}{dx} = \frac{\log(\cos y) + y \tan x} {\log(\cos x) + x \tan y}

\displaystyle \textbf{Question 57: }~\text{If }\cos y=x\cos(a+y),\text{ where }\cos a\neq\pm1,\text{ prove that }\frac{dy}{dx}=\frac{\cos^2(a+y)}{\sin a}\;[\text{CBSE 2014}]
\displaystyle \text{Answer:}

\displaystyle \cos y = x \cos(a + y)

\displaystyle \Rightarrow -\sin y \frac{dy}{dx} = \cos(a + y) - x \sin(a + y)\frac{dy}{dx}

\displaystyle \Rightarrow -\sin y \frac{dy}{dx} + x \sin(a + y)\frac{dy}{dx} = \cos(a + y)

\displaystyle \Rightarrow \frac{dy}{dx} \left[x \sin(a + y) - \sin y\right] = \cos(a + y)

\displaystyle \Rightarrow \frac{dy}{dx} \left[\frac{\cos y}{\cos(a + y)}\sin(a + y) - \sin y\right] = \cos(a + y)

\displaystyle \Rightarrow \frac{dy}{dx} \left[\frac{\cos y \sin(a + y) - \sin y \cos(a + y)} {\cos(a + y)}\right] = \cos(a + y)

\displaystyle \Rightarrow \frac{dy}{dx} \left[\frac{\sin(a + y - y)}{\cos(a + y)}\right] = \cos(a + y)

\displaystyle \Rightarrow \frac{dy}{dx} \left[\frac{\sin a}{\cos(a + y)}\right] = \cos(a + y)

\displaystyle \therefore \frac{dy}{dx} = \frac{\cos^{2}(a + y)}{\sin a}

\displaystyle \textbf{Question 58: }~\text{If }(x-y)e^{\frac{x}{x-y}}=a,\text{ prove that }y\frac{dy}{dx}+x=2y\;[\text{CBSE 2014}]
\displaystyle \text{Answer:}

\displaystyle (x - y)\,e^{\frac{x}{x - y}} = a

\displaystyle \text{Taking log on both sides, we get}

\displaystyle \log(x - y) + \frac{x}{x - y} = \log a

\displaystyle \Rightarrow \frac{1}{x - y}\left(1 - \frac{dy}{dx}\right) + \frac{(x - y) - x\left(1 - \frac{dy}{dx}\right)}{(x - y)^{2}} = 0

\displaystyle \Rightarrow \frac{1 - \frac{dy}{dx}}{x - y} + \frac{x\frac{dy}{dx} - y}{(x - y)^{2}} = 0

\displaystyle \Rightarrow \frac{(x - y)\left(1 - \frac{dy}{dx}\right) + x\frac{dy}{dx} - y}{(x - y)^{2}} = 0

\displaystyle \Rightarrow x - x\frac{dy}{dx} - y + y\frac{dy}{dx} + x\frac{dy}{dx} - y = 0

\displaystyle \Rightarrow x - 2y + y\frac{dy}{dx} = 0

\displaystyle \therefore y\frac{dy}{dx} + x = 2y

\displaystyle \textbf{Question 59: }~\text{If }x=e^{x/y},\text{ prove that }\frac{dy}{dx}=\frac{x-y}{x\log x}\;
\displaystyle \text{Answer:}

\displaystyle x = e^{\frac{x}{y}}

\displaystyle \text{Taking logarithm on both sides, we get}

\displaystyle \log x = \frac{x}{y}

\displaystyle \Rightarrow y \log x = x

\displaystyle \Rightarrow \log x \frac{dy}{dx} + y \frac{1}{x} = 1

\displaystyle \Rightarrow \log x \frac{dy}{dx} = 1 - \frac{y}{x}

\displaystyle \Rightarrow \log x \frac{dy}{dx} = \frac{x - y}{x}

\displaystyle \Rightarrow x \log x \frac{dy}{dx} = x - y

\displaystyle \therefore \frac{dy}{dx} = \frac{x - y}{x \log x}

\displaystyle \textbf{Question 60: }~\text{If }y=x^{\tan x}+\sqrt{\frac{x^2+1}{2}},\text{ find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle y = x^{\tan x} + \sqrt{\frac{x^{2} + 1}{2}}

\displaystyle \text{Differentiate both sides with respect to } x

\displaystyle \text{For } x^{\tan x}, \text{ take log differentiation:}

\displaystyle \log\left(x^{\tan x}\right) = \tan x \log x

\displaystyle \Rightarrow \frac{1}{x^{\tan x}} \frac{d}{dx}\left(x^{\tan x}\right) = \sec^{2}x \log x + \frac{\tan x}{x}

\displaystyle \Rightarrow \frac{d}{dx}\left(x^{\tan x}\right) = x^{\tan x}\left(\sec^{2}x \log x + \frac{\tan x}{x}\right)

\displaystyle \text{Now, } \frac{d}{dx}\left(\sqrt{\frac{x^{2}+1}{2}}\right) = \frac{x}{\sqrt{2x^{2}+2}}

\displaystyle \therefore \frac{dy}{dx} = x^{\tan x}\left(\frac{\tan x}{x} + \sec^{2}x \log x\right) + \frac{x}{\sqrt{2x^{2}+2}}

\displaystyle \textbf{Question 61: }~\text{If }y=1+\frac{\alpha}{\left(\frac{1}{x}-\alpha\right)}+\frac{\beta/x}{\left(\frac{1}{x}-\alpha\right)\left(\frac{1}{x}-\beta\right)}+\frac{\gamma/x^2}{\left(\frac{1}{x}-\alpha\right)\left(\frac{1}{x}-\beta\right)\left(\frac{1}{x}-\gamma\right)},\text{ find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle y = \frac{\frac{1}{x^{3}}} {\left(\frac{1}{x}-\alpha\right) \left(\frac{1}{x}-\beta\right) \left(\frac{1}{x}-\gamma\right)}

\displaystyle \text{Taking log on both sides, we get}

\displaystyle \log y = -3\log x - \log\left(\frac{1}{x}-\alpha\right) - \log\left(\frac{1}{x}-\beta\right) - \log\left(\frac{1}{x}-\gamma\right)

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = -\frac{3}{x} - \frac{1}{\frac{1}{x}-\alpha}\left(-\frac{1}{x^{2}}\right) - \frac{1}{\frac{1}{x}-\beta}\left(-\frac{1}{x^{2}}\right) - \frac{1}{\frac{1}{x}-\gamma}\left(-\frac{1}{x^{2}}\right)

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = -\frac{3}{x} + \frac{1}{x^{2}}\left[ \frac{1}{\frac{1}{x}-\alpha} + \frac{1}{\frac{1}{x}-\beta} + \frac{1}{\frac{1}{x}-\gamma} \right]

\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{1}{x}\left[ \frac{\alpha}{\frac{1}{x}-\alpha} + \frac{\beta}{\frac{1}{x}-\beta} + \frac{\gamma}{\frac{1}{x}-\gamma} \right]

\displaystyle \therefore \frac{dy}{dx} = \frac{y}{x}\left( \frac{\alpha}{\frac{1}{x}-\alpha} + \frac{\beta}{\frac{1}{x}-\beta} + \frac{\gamma}{\frac{1}{x}-\gamma} \right)

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