Differentiate the following functions with respect to x :

\displaystyle \textbf{Question 1: }x^4-2\sin x+3\cos x
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(x^4-2\sin x+3\cos x)
\displaystyle =\frac{d}{dx}(x^4)-2\frac{d}{dx}(\sin x)+3\frac{d}{dx}(\cos x)
\displaystyle =4x^3-2\cos x-3\sin x
\displaystyle \\

\displaystyle \textbf{Question 2: }3^x+x^3+3^3
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(3^x+x^3+3^3)
\displaystyle =\frac{d}{dx}(3^x)+\frac{d}{dx}(x^3)+\frac{d}{dx}(3^3)
\displaystyle =3^x\log_e3+3x^2
\displaystyle \\

\displaystyle \textbf{Question 3: }\frac{x^3}{3}-2\sqrt{x}+\frac{5}{x^2}
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left(\frac{x^3}{3}-2\sqrt{x}+\frac{5}{x^2}\right)
\displaystyle =\frac{d}{dx}\left(\frac{x^3}{3}\right)-2\frac{d}{dx}(\sqrt{x})+\frac{d}{dx}\left(\frac{5}{x^2}\right)
\displaystyle =\frac{1}{3}(3x^2)-2\cdot\frac{1}{2}x^{-\frac{1}{2}}+5(-2)x^{-3}
\displaystyle =x^2-x^{-\frac{1}{2}}-10x^{-3}
\displaystyle \\

\displaystyle \textbf{Question 4: }e^{x\log a}+e^{a\log x}+e^{a\log a}
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left(e^{x\log a}+e^{a\log x}+e^{a\log a}\right)
\displaystyle =\frac{d}{dx}(e^{x\log a})+\frac{d}{dx}(e^{a\log x})+\frac{d}{dx}(e^{a\log a})
\displaystyle =\frac{d}{dx}(a^x)+\frac{d}{dx}(x^a)+\frac{d}{dx}(a^a)
\displaystyle =a^x\log_ea+ax^{a-1}
\displaystyle \\

\displaystyle \textbf{Question 5: }(2x^2+1)(3x+2)
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left((2x^2+1)(3x+2)\right)
\displaystyle =\frac{d}{dx}\left(6x^3+4x^2+3x+2\right)
\displaystyle =\frac{d}{dx}(6x^3)+\frac{d}{dx}(4x^2)+\frac{d}{dx}(3x)+\frac{d}{dx}(2)
\displaystyle =18x^2+8x+3
\displaystyle \\

\displaystyle \textbf{Question 6: }\log_3x+3\log_ex+2\tan x
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left(\log_3x+3\log_ex+2\tan x\right)
\displaystyle =\frac{d}{dx}(\log_3x)+3\frac{d}{dx}(\log_ex)+2\frac{d}{dx}(\tan x)
\displaystyle =\frac{d}{dx}\left(\frac{\log_ex}{\log_e3}\right)+3\frac{d}{dx}(\log_ex)+2\frac{d}{dx}(\tan x)
\displaystyle =\frac{1}{\log_e3}\cdot\frac{1}{x}+\frac{3}{x}+2\sec^2x
\displaystyle =\frac{1}{x\log_e3}+\frac{3}{x}+2\sec^2x
\displaystyle \\

\displaystyle \textbf{Question 7: }\left(x+\frac{1}{x}\right)\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left[\left(x+\frac{1}{x}\right)\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\right]
\displaystyle =\frac{d}{dx}\left[(x+x^{-1})(x^{\frac12}+x^{-\frac12})\right]
\displaystyle =\frac{d}{dx}\left(x^{\frac32}+x^{\frac12}+x^{-\frac12}+x^{-\frac32}\right)
\displaystyle =\frac{d}{dx}\left(x^{\frac32}\right)+\frac{d}{dx}\left(x^{\frac12}\right)+\frac{d}{dx}\left(x^{-\frac12}\right)+\frac{d}{dx}\left(x^{-\frac32}\right)
\displaystyle =\frac32x^{\frac12}+\frac12x^{-\frac12}-\frac12x^{-\frac32}-\frac32x^{-\frac52}
\displaystyle \\

\displaystyle \textbf{Question 8: }\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^3
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left[\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^3\right]
\displaystyle =\frac{d}{dx}\left[\left(\sqrt{x}\right)^3+3\left(\sqrt{x}\right)^2\left(\frac{1}{\sqrt{x}}\right)+3\left(\sqrt{x}\right)\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^3\right]
\displaystyle =\frac{d}{dx}\left(x^{\frac32}\right)+3\frac{d}{dx}\left(x^{\frac12}\right)+3\frac{d}{dx}\left(x^{-\frac12}\right)+\frac{d}{dx}\left(x^{-\frac32}\right)
\displaystyle =\frac32x^{\frac12}+3\cdot\frac12x^{-\frac12}+3\cdot\left(-\frac12\right)x^{-\frac32}-\frac32x^{-\frac52}
\displaystyle =\frac32x^{\frac12}+\frac32x^{-\frac12}-\frac32x^{-\frac32}-\frac32x^{-\frac52}
\displaystyle \\

\displaystyle \textbf{Question 9: }\frac{2x^2+3x+4}{x}
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left(\frac{2x^2+3x+4}{x}\right)
\displaystyle =\frac{d}{dx}\left(\frac{2x^2}{x}\right)+\frac{d}{dx}\left(\frac{3x}{x}\right)+\frac{d}{dx}\left(\frac{4}{x}\right)
\displaystyle =2\frac{d}{dx}(x)+3\frac{d}{dx}(1)+4\frac{d}{dx}(x^{-1})
\displaystyle =2+3(0)+4(-1)x^{-2}
\displaystyle =2-\frac{4}{x^2}
\displaystyle \\

\displaystyle \textbf{Question 10: }\frac{(x^3+1)(x-2)}{x^2}
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left(\frac{(x^3+1)(x-2)}{x^2}\right)
\displaystyle =\frac{d}{dx}\left(\frac{x^4-2x^3+x-2}{x^2}\right)
\displaystyle =\frac{d}{dx}(x^2)-2\frac{d}{dx}(x)+\frac{d}{dx}(x^{-1})-2\frac{d}{dx}(x^{-2})
\displaystyle =2x-2-x^{-2}-2(-2)x^{-3}
\displaystyle =2x-2-\frac{1}{x^2}+\frac{4}{x^3}
\displaystyle \\

\displaystyle \textbf{Question 11: }\frac{a\cos x+b\sin x+c}{\sin x}
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left(\frac{a\cos x+b\sin x+c}{\sin x}\right)
\displaystyle =\frac{d}{dx}\left(\frac{a\cos x}{\sin x}\right)+\frac{d}{dx}\left(\frac{b\sin x}{\sin x}\right)+\frac{d}{dx}\left(\frac{c}{\sin x}\right)
\displaystyle =a\frac{d}{dx}(\cot x)+\frac{d}{dx}(b)+c\frac{d}{dx}(\mathrm{cosec}\,x)
\displaystyle =-a\,\mathrm{cosec}^2x+0-c\,\mathrm{cosec}\,x\cot x
\displaystyle =-a\,\mathrm{cosec}^2x-c\,\mathrm{cosec}\,x\cot x
\displaystyle \\

\displaystyle \textbf{Question 12: }2\sec x+3\cot x-4\tan x
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(2\sec x+3\cot x-4\tan x)
\displaystyle =2\frac{d}{dx}(\sec x)+3\frac{d}{dx}(\cot x)-4\frac{d}{dx}(\tan x)
\displaystyle =2\sec x\tan x-3\mathrm{cosec}^2x-4\sec^2x
\displaystyle \\

\displaystyle \textbf{Question 13: }a_0x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_{n-1}x+a_n
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left(a_0x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_{n-1}x+a_n\right)
\displaystyle =a_0\frac{d}{dx}(x^n)+a_1\frac{d}{dx}(x^{n-1})+a_2\frac{d}{dx}(x^{n-2})+\cdots+a_{n-1}\frac{d}{dx}(x)+a_n\frac{d}{dx}(1)
\displaystyle =na_0x^{n-1}+(n-1)a_1x^{n-2}+(n-2)a_2x^{n-3}+\cdots+a_{n-1}
\displaystyle \\

\displaystyle \textbf{Question 14: }\frac{1}{\sin x}+2^{x+3}+\frac{4}{\log_x3}
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left(\frac{1}{\sin x}+2^{x+3}+\frac{4}{\log_x3}\right)
\displaystyle =\frac{d}{dx}\left(\mathrm{cosec}\,x+2^3\cdot2^x+\frac{4}{\frac{\log_e3}{\log_ex}}\right)
\displaystyle =\frac{d}{dx}(\mathrm{cosec}\,x)+2^3\frac{d}{dx}(2^x)+\frac{4}{\log_e3}\frac{d}{dx}(\log_ex)
\displaystyle =-\mathrm{cosec}\,x\cot x+2^3\cdot2^x\log_e2+\frac{4}{\log_e3}\cdot\frac{1}{x}
\displaystyle =-\mathrm{cosec}\,x\cot x+2^{x+3}\log_e2+\frac{4}{x\log_e3}
\displaystyle \\

\displaystyle \textbf{Question 15: }\frac{(x+5)(2x^2-1)}{x}
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left(\frac{(x+5)(2x^2-1)}{x}\right)
\displaystyle =\frac{d}{dx}\left(\frac{2x^3+10x^2-x-5}{x}\right)
\displaystyle =\frac{d}{dx}(2x^2)+\frac{d}{dx}(10x)-\frac{d}{dx}(1)-\frac{d}{dx}(5x^{-1})
\displaystyle =4x+10+5x^{-2}
\displaystyle =4x+10+\frac{5}{x^2}
\displaystyle \\

\displaystyle \textbf{Question 16: }\log\left(\frac{1}{\sqrt{x}}\right)+5x^a-3a^x+\sqrt[3]{x^2}+6\sqrt[4]{x^{-3}}
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left[\log\left(\frac{1}{\sqrt{x}}\right)+5x^a-3a^x+\sqrt[3]{x^2}+6\sqrt[4]{x^{-3}}\right]
\displaystyle =\frac{d}{dx}\left(\log_e x^{-\frac12}\right)+5\frac{d}{dx}(x^a)-3\frac{d}{dx}(a^x)+\frac{d}{dx}\left(x^{\frac23}\right)+6\frac{d}{dx}\left(x^{-\frac34}\right)
\displaystyle =\frac{d}{dx}\left(-\frac12\log_ex\right)+5\frac{d}{dx}(x^a)-3\frac{d}{dx}(a^x)+\frac{d}{dx}\left(x^{\frac23}\right)+6\frac{d}{dx}\left(x^{-\frac34}\right)
\displaystyle =-\frac{1}{2}\cdot\frac{1}{x}+5ax^{a-1}-3a^x\log_ea+\frac23x^{-\frac13}+6\left(-\frac34\right)x^{-\frac74}
\displaystyle =-\frac{1}{2x}+5ax^{a-1}-3a^x\log_ea+\frac23x^{-\frac13}-\frac92x^{-\frac74}
\displaystyle \\

\displaystyle \textbf{Question 17: }\cos(x+a)
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\big(\cos(x+a)\big)
\displaystyle =\frac{d}{dx}\big(\cos x\cos a-\sin x\sin a\big)
\displaystyle =\cos a\frac{d}{dx}(\cos x)-\sin a\frac{d}{dx}(\sin x)
\displaystyle =-\cos a\sin x-\sin a\cos x
\displaystyle =-(\sin x\cos a+\cos x\sin a)
\displaystyle =-\sin(x+a)
\displaystyle \\

\displaystyle \textbf{Question 18: }\frac{\cos(x-2)}{\sin x}
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left(\frac{\cos(x-2)}{\sin x}\right)
\displaystyle =\frac{d}{dx}\left(\frac{\cos x\cos2+\sin x\sin2}{\sin x}\right)
\displaystyle =\frac{d}{dx}\left(\cos2\cot x+\sin2\right)
\displaystyle =\cos2\frac{d}{dx}(\cot x)+\frac{d}{dx}(\sin2)
\displaystyle =-\cos2\,\mathrm{cosec}^2x+0
\displaystyle =-\cos2\,\mathrm{cosec}^2x
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }y=\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)^2,\text{ find }\frac{dy}{dx}
\displaystyle \text{at }x=\frac{\pi}{4}.
\displaystyle \text{Answer:}
\displaystyle y=\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)^2
\displaystyle =\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}
\displaystyle =1+\sin x
\displaystyle \therefore \frac{dy}{dx}=\frac{d}{dx}(1+\sin x)
\displaystyle =\cos x
\displaystyle \left.\frac{dy}{dx}\right|_{x=\frac{\pi}{4}}=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }y=\frac{2-3\cos x}{\sin x},\text{ find }\frac{dy}{dx}\text{ at }x=\frac{\pi}{4}.
\displaystyle \text{Answer:}
\displaystyle y=\frac{2-3\cos x}{\sin x}
\displaystyle =2\,\mathrm{cosec}\,x-3\cot x
\displaystyle \frac{dy}{dx}=2\frac{d}{dx}(\mathrm{cosec}\,x)-3\frac{d}{dx}(\cot x)
\displaystyle =-2\,\mathrm{cosec}\,x\cot x+3\,\mathrm{cosec}^2x
\displaystyle \left.\frac{dy}{dx}\right|_{x=\frac{\pi}{4}}=-2\,\mathrm{cosec}\frac{\pi}{4}\cot\frac{\pi}{4}+3\,\mathrm{cosec}^2\frac{\pi}{4}
\displaystyle =-2(\sqrt{2})(1)+3(2)
\displaystyle =6-2\sqrt{2}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Find the slope of the tangent to the curve }f(x)=2x^6+x^4-1
\displaystyle \text{at }x=1.
\displaystyle \text{Answer:}
\displaystyle f'(x)=\frac{d}{dx}(2x^6+x^4-1)
\displaystyle =2\frac{d}{dx}(x^6)+\frac{d}{dx}(x^4)-\frac{d}{dx}(1)
\displaystyle =12x^5+4x^3
\displaystyle \therefore f'(1)=12(1)^5+4(1)^3
\displaystyle =12+4=16
\displaystyle \therefore \text{The slope of the tangent is }16.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }y=\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}},\text{ prove that }2xy\frac{dy}{dx}
\displaystyle =\frac{x}{a}-\frac{a}{x}.
\displaystyle \text{Answer:}
\displaystyle y=\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}
\displaystyle =\frac{1}{\sqrt{a}}\sqrt{x}+\sqrt{a}\,x^{-\frac12}
\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{a}}\frac{d}{dx}(\sqrt{x})+\sqrt{a}\frac{d}{dx}\left(x^{-\frac12}\right)
\displaystyle =\frac{1}{2\sqrt{a}\sqrt{x}}-\frac{\sqrt{a}}{2x\sqrt{x}}
\displaystyle =\frac{1}{2x}\left(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}}\right)
\displaystyle \therefore 2x\frac{dy}{dx}=\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}}
\displaystyle \text{Multiplying both sides by }y,\text{ we get}
\displaystyle 2xy\frac{dy}{dx}=\left(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}}\right)\left(\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}\right)
\displaystyle =\frac{x}{a}-\frac{a}{x}
\displaystyle \therefore 2xy\frac{dy}{dx}=\frac{x}{a}-\frac{a}{x}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Find the rate at which the function }f(x)=x^4-2x^3+3x^2+x+5
\displaystyle \text{changes with respect to }x.
\displaystyle \text{Answer:}
\displaystyle \text{Rate}=f'(x)
\displaystyle =\frac{d}{dx}(x^4-2x^3+3x^2+x+5)
\displaystyle =\frac{d}{dx}(x^4)-2\frac{d}{dx}(x^3)+3\frac{d}{dx}(x^2)+\frac{d}{dx}(x)+\frac{d}{dx}(5)
\displaystyle =4x^3-6x^2+6x+1
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }y=\frac{2x^9}{3}-\frac{5}{7}x^7+6x^3-x,\text{ find }\frac{dy}{dx}
\displaystyle \text{at }x=1.
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(\frac{2x^9}{3}-\frac{5}{7}x^7+6x^3-x\right)
\displaystyle =\frac{2}{3}\frac{d}{dx}(x^9)-\frac{5}{7}\frac{d}{dx}(x^7)+6\frac{d}{dx}(x^3)-\frac{d}{dx}(x)
\displaystyle =\frac{2}{3}(9x^8)-\frac{5}{7}(7x^6)+6(3x^2)-1
\displaystyle =6x^8-5x^6+18x^2-1
\displaystyle \left.\frac{dy}{dx}\right|_{x=1}=6(1)^8-5(1)^6+18(1)^2-1
\displaystyle =6-5+18-1=18
\displaystyle \therefore \left.\frac{dy}{dx}\right|_{x=1}=18.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If for }f(x)=\lambda x^2+\mu x+12,\ f'(4)=15\text{ and }f'(2)=11,\text{ find }\lambda\text{ and }\mu.
\displaystyle \text{Answer:}
\displaystyle f'(x)=\frac{d}{dx}(\lambda x^2+\mu x+12)
\displaystyle =\lambda\frac{d}{dx}(x^2)+\mu\frac{d}{dx}(x)+\frac{d}{dx}(12)
\displaystyle =2\lambda x+\mu
\displaystyle \text{Given }f'(4)=15
\displaystyle \Rightarrow 2\lambda(4)+\mu=15
\displaystyle \Rightarrow 8\lambda+\mu=15\hspace{0.5cm}\text{... ... ... ... ... (i)}
\displaystyle \text{Also }f'(2)=11
\displaystyle \Rightarrow 2\lambda(2)+\mu=11
\displaystyle \Rightarrow 4\lambda+\mu=11\hspace{0.5cm}\text{... ... ... ... ... (ii)}
\displaystyle \text{Subtracting (ii) from (i),}
\displaystyle 4\lambda=4
\displaystyle \therefore \lambda=1
\displaystyle \text{Substituting in (ii),}
\displaystyle 4(1)+\mu=11
\displaystyle \therefore \mu=7
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{For the function }f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\cdots+\frac{x^2}{2}+x+1,
\displaystyle \text{prove that }f'(1)=100f'(0).
\displaystyle \text{Answer:}
\displaystyle f'(x)=\frac{d}{dx}\left(\frac{x^{100}}{100}+\frac{x^{99}}{99}+\cdots+\frac{x^2}{2}+x+1\right)
\displaystyle =x^{99}+x^{98}+\cdots+x+1
\displaystyle \therefore f'(1)=1^{99}+1^{98}+\cdots+1+1=100
\displaystyle \therefore f'(0)=0^{99}+0^{98}+\cdots+0+1=1
\displaystyle \therefore f'(1)=100=100f'(0)
\displaystyle \\

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