\displaystyle \textbf{Question 1: }~\text{If }y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots\ \text{to }\infty}}},\\ \text{ prove that }\frac{dy}{dx}=\frac{1}{2y-1}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } y = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}

\displaystyle \Rightarrow y = \sqrt{x + y}

\displaystyle \text{Squaring both sides, we get}

\displaystyle y^{2} = x + y

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

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

\displaystyle \therefore \frac{dy}{dx} = \frac{1}{2y - 1}

\displaystyle \textbf{Question 2: }~\text{If }y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\cdots\ \text{to }\infty}}},\\ \text{ prove that }\frac{dy}{dx}=\frac{\sin x}{1-2y}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } y = \sqrt{\cos x + \sqrt{\cos x + \sqrt{\cos x + \cdots}}}

\displaystyle \Rightarrow y = \sqrt{\cos x + y}

\displaystyle \text{Squaring both sides, we get}

\displaystyle y^{2} = \cos x + y

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

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

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{-\sin x}{2y - 1}

\displaystyle \therefore \frac{dy}{dx} = \frac{\sin x}{1 - 2y}

\displaystyle \textbf{Question 3: }~\text{If }y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\cdots\ \text{to }\infty}}},\\ \text{ prove that }(2y-1)\frac{dy}{dx}=\frac{1}{x}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } y = \sqrt{\log x + \sqrt{\log x + \sqrt{\log x + \cdots}}}

\displaystyle \Rightarrow y = \sqrt{\log x + y}

\displaystyle \text{Squaring both sides, we get}

\displaystyle y^{2} = \log x + y

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

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

\displaystyle \therefore \frac{dy}{dx} = \frac{1}{x(2y - 1)}

\displaystyle \textbf{Question 4: }~\text{If }y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\cdots\ \text{to }\infty}}},\\ \text{ prove that }\frac{dy}{dx}=\frac{\sec^2 x}{2y-1}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } y = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + \cdots}}}

\displaystyle \Rightarrow y = \sqrt{\tan x + y}

\displaystyle \text{Squaring both sides, we get}

\displaystyle y^{2} = \tan x + y

\displaystyle \Rightarrow 2y \frac{dy}{dx} = \sec^{2}x + \frac{dy}{dx}

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

\displaystyle \therefore \frac{dy}{dx} = \frac{\sec^{2}x}{2y - 1}

\displaystyle \textbf{Question 5: }~\text{If }y=(\sin x)^{(\sin x)^{(\sin x)^{\cdots\ \infty}}},\text{ prove that }\frac{dy}{dx}=\frac{y^2\cot x}{1-y\log(\sin x)}
\displaystyle \text{Answer:}

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

\displaystyle \Rightarrow y = (\sin x)^{y}

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

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

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

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

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

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

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

\displaystyle \therefore \frac{dy}{dx} = \frac{y^{2}\cot x}{1 - y\log(\sin x)}

\displaystyle \textbf{Question 6: }~\text{If }y=(\tan x)^{(\tan x)^{(\tan x)^{\cdots\ \infty}}},\text{ prove that }\frac{dy}{dx}=2\text{ at }x=\frac{\pi}{4}
\displaystyle \text{Answer:}

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

\displaystyle \Rightarrow y = (\tan x)^{y}

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

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

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

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

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

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

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

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

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

\displaystyle \text{Now, at } x=\frac{\pi}{4}:

\displaystyle \tan\frac{\pi}{4}=1,\quad \sec^{2}\frac{\pi}{4}=2,\quad y=(\tan\tfrac{\pi}{4})^{(\tan\tfrac{\pi}{4})^{\cdots}}=1

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}} = \frac{1^{2}\cdot 2}{1\,(1-1\cdot\log 1)}

\displaystyle \therefore \left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}} = 2

\displaystyle \textbf{Question 7: }~\text{If }y=e^{x^{e^x}}+x^{e^x}+e^{x^e},\text{ prove that } \\ \frac{dy}{dx}=e^{x^{e^x}}\cdot x^{e^x}\!\left(\frac{1}{x}+e^x\log x\right)+x^{e^e}\cdot e^x\!\left(\frac{1}{x}+e^x\log x\right)+e^{x^e}\cdot x^{e-1}(1+e\log x)
\displaystyle \text{Answer:}

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

\displaystyle \Rightarrow y = u + v + w

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

\displaystyle \text{where } u = e^{x^{e^{x}}},\ v = x^{e^{x}} \text{ and } w = e^{x^{x^{e}}}

\displaystyle \text{Now, } u = e^{x^{e^{x}}} \quad \text{...(ii)}

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

\displaystyle \log u = \log e^{x^{e^{x}}}

\displaystyle \Rightarrow \log u = x^{e^{x}} \log e

\displaystyle \Rightarrow \log u = x^{e^{x}} \quad \text{...(iii)}

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

\displaystyle \log \log u = \log x^{e^{x}}

\displaystyle \Rightarrow \log \log u = e^{x} \log x

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

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

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

\displaystyle \Rightarrow \frac{du}{dx} = u \log u \left[\frac{e^{x}}{x} + e^{x}\log x\right]

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

\displaystyle \text{[Using equation (ii) and (iii)]}

\displaystyle \text{Now, } v = x^{e^{e^{x}}} \quad \text{...(iv)}

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

\displaystyle \log v = \log x^{e^{e^{x}}}

\displaystyle \Rightarrow \log v = e^{e^{x}}\log x

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

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} = e^{e^{x}}\left(\frac{1}{x}\right) + \log x\, e^{e^{x}}\frac{d}{dx}(e^{x})

\displaystyle \Rightarrow \frac{dv}{dx} = v\left[e^{e^{x}}\left(\frac{1}{x}\right) + \log x\, e^{e^{x}}e^{x}\right]

\displaystyle \Rightarrow \frac{dv}{dx} = x^{e^{e^{x}}}\times e^{e^{x}} \left[\frac{1}{x} + e^{x}\log x\right] \quad \text{...(B)}

\displaystyle \{\ \text{Using equation (4)}\ \}

\displaystyle \text{Now, } w = e^{x^{x^{e}}} \quad \text{...(v)}

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

\displaystyle \log w = \log e^{x^{x^{e}}}

\displaystyle \Rightarrow \log w = x^{x^{e}}\log e

\displaystyle \Rightarrow \log w = x^{x^{e}} \quad \text{...(vi)}

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

\displaystyle \log \log w = \log x^{x^{e}}

\displaystyle \Rightarrow \log \log w = x^{e}\log x

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

\displaystyle \Rightarrow \frac{1}{\log w}\left(\frac{1}{w}\right)\frac{dw}{dx} = x^{e}\left(\frac{1}{x}\right) + \log x\, e x^{e-1}

\displaystyle \Rightarrow \frac{dw}{dx} = w \log w \left[x^{e-1} + e\log x\, x^{e-1}\right]

\displaystyle \Rightarrow \frac{dw}{dx} = e^{x^{x^{e}}}\,x^{x^{e}}\,x^{e-1}(1 + e\log x) \;-\;-\;-\;-\; (C)

\displaystyle \text{[using equation (v), (vi)]}

\displaystyle \text{Using equation (A), (B) and (C) in equation (i), we get}

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

\displaystyle \textbf{Question 8: }~\text{If }y=(\cos x)^{(\cos x)^{(\cos x)^{\cdots\ \infty}}},\text{ prove that }\frac{dy}{dx}=-\frac{y^2\tan x}{1-y\log(\cos x)}\;
\displaystyle \text{Answer:}

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

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

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

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

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

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

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

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

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

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

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{y^{2}\tan x}{1 - y \log \cos x}

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