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Also Read - RD Sharma Solution for Class 9 to 12 Maths
The RD Sharma class 12 chapter 10 exercise MCQ is one of the best NCERT solutions trusted and recommended by hundreds of students in their exams. It can be used for both school and board exams. The 10th chapter of the maths book deals with Differentiations in trigonometric equations. It will teach students about Logarithmic differentiation, differentiation of inverse trigonometric functions, Differentiation of a function. Exercise MCQ includes 33 questions on all the concepts that students have learned in this chapter.
RD Sharma Class 12 Solutions Chapter 10 MCQ Differentiation - Other Exercise
Differentiation Excercise: MCQ
Differentiation exercise Multiple choice question 1
Answer: $\frac{1}{2e}$Hint: Use quotient rule to differentiate the function
Given: $f(x)=\log _{x^{2}}\left(\log _{e}\right) x$Solution: $f(x)=\log _{x^{2}}\left(\log _{e}\right) x$$=\frac{\log (\log x)}{(\log (x))^{2}}$$f(x)=\frac{\log (\log x)}{2 \log x}$Use quotient rule
$f^{\prime}(x)=\frac{\frac{1}{2}\left[\log \left(\frac{1}{\log x}\right)\left(\frac{1}{x}\right)-\log (\log (x))\left(\frac{1}{x}\right)\right]}{(\log x)^{2}}$$=\frac{1}{2} \frac{\left[\frac{(1-\log (\log x))}{x}\right]}{(\log x)^{2}}$Put x=e
$f^{\prime}(e)=\left(\frac{1}{2}\right) \frac{\left[\frac{(1-\log (\log e))}{e}\right]}{(\log e)^{2}}$$\begin{aligned} &=\left(\frac{1}{2}\right) \frac{(1-0)}{e} \quad[\text { Since } \log e=1] \\\\ &f^{\prime}(e)=\frac{1}{2 e} \end{aligned}$Differentiation exercise Multiple choice question 2
Answer:$(x \log )^{-1}$Hint: Use differential formula
Given:$f(x)=\log x$Solution: We have
$\begin{aligned} &f(x)=\log x \\\\ &f(\log x)=\log (\log x) \\\\ &f^{\prime}(\log x)=\frac{1}{\log x} \frac{d}{d x}(\log x) \end{aligned}$$=\frac{1}{x \log x}$$f^{\prime}(\log x)=(x \log x)^{-1}$Differentiation exercise Multiple choice question 3
Answer:$\left(\frac{2}{3}\right)^{\frac{1}{2}}$Hint: Differentiate the given function and replace x by
$\frac{\pi }{6}$ and solve
Given:$\cot ^{-1}\left\{(\cos 2 x)^{\frac{1}{2}}\right\} \text { at } x=\frac{\pi}{6}$Solution: $\begin{aligned} &y=\cot ^{-1}(\sqrt{\cos 2 x}) \\\\ &\frac{d y}{d x}=\frac{-1}{1+\cos 2 x} \frac{d}{d x} \sqrt{\cos 2 x} \end{aligned}$$\frac{d y}{d x}=\left[\frac{-1}{1+\cos 2 x}\right]\left[\frac{1}{2}(\sqrt{\cos 2 x})\right](-2 \sin 2 x)$$=\frac{\sin 2 x}{(1+\cos 2 x) \sqrt{\cos 2 x}}$$\text { At } x=\frac{\pi}{6}$$\frac{d y}{d x}=\frac{\sin \frac{\pi}{3}}{\left(1+\cos \frac{\pi}{3}\right) \sqrt{\cos \frac{\pi}{3}}}$$=\frac{\left(\frac{\sqrt{3}}{2}\right)}{\left(1+\frac{1}{2}\right) \sqrt{\frac{1}{2}}}$$\begin{aligned} &=\sqrt{\frac{2}{3}} \\\\ &=\left(\frac{2}{3}\right)^{\frac{1}{2}} \end{aligned}$Differentiation exercise Multiple choice question 4
Answer: $\frac{x}{\sqrt{1+x^{2}}}$Hint: Differentiate the function w.r.t x
Given:$\sec \left(\tan ^{-1} x\right)$Solution: $\begin{aligned} &y=\sec \left(\tan ^{-1} x\right) \\\\ &\frac{d y}{d x}=\sec \left(\tan ^{-1} x\right) \tan \left(\tan ^{-1} x\right) \times \frac{d}{d x}\left(\tan ^{-1} x\right) \end{aligned}$$\begin{aligned} &=\sec \left(\tan ^{-1} x\right) \tan \left(\tan ^{-1} x\right) \times \frac{1}{\sqrt{1+x^{2}}} \\\\ &=y \tan \left(\tan ^{-1} x\right) \times \frac{1}{\sqrt{1+x^{2}}} \end{aligned}$$=y\left(\frac{x}{\sqrt{1+x^{2}}}\right) \; \; \; \; \; \; \quad\left[\tan \left(\tan ^{-1} x\right)=x\right]$This is the equation of differential equation which have co-efficient=
$\left(\frac{x}{\sqrt{1+x^{2}}}\right)$Differentiation exercise Multiple choice question 5
Answer:$\frac{1}{2}$Hint: Differentiate the function and replace x by
$\frac{\pi }{6}$ solve
Given:$f(x)=\tan ^{-1} \sqrt{\frac{1+\sin x}{1-\sin x}}, 0 \leq x \leq \frac{\pi}{2}, f^{\prime}\left(\frac{\pi}{6}\right)$Solution: $y=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$$=\tan ^{-1}\left(\sqrt{\frac{1-\cos \left(\frac{\pi}{2}+x\right)}{1+\cos \left(\frac{\pi}{2}+x\right)}}\right)$$=\tan ^{-1}\left(\sqrt{\frac{2 \sin ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \cos ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}}\right)$$y=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right)=\frac{\pi}{4}+\frac{x}{2} \; \; \; \; \; \; \; \quad\left[\tan ^{-1}(\tan x)=x\right]$$\frac{d y}{d x}=\frac{1}{2}$Differentiation exercise Multiple choice question 6
Answer:$\left(1+\frac{1}{x}\right)^{x}\left(1+\frac{1}{x}\right)-\frac{1}{x+1}$Hint: Differentiate the function w.r.t x
Given:$y=\left(1+\frac{1}{x}\right)^{x}$Solution: Let
$y=\left(1+\frac{1}{x}\right)^{x}$Taking log on both sides
$\begin{aligned} &\log y=x \log \left(1+\frac{1}{x}\right) \\\\ &\frac{1}{y} \frac{d y}{d x}=x \frac{d}{d x} \log \left(1+\frac{1}{x}\right)+\log \left(1+\frac{1}{x}\right) \frac{d}{d x}(x) \end{aligned}$$=x\left(\frac{1}{1+\frac{1}{x}}\right) \frac{d}{d x}\left(1+\frac{1}{x}\right)+\log \left(1+\frac{1}{x}\right)$$\begin{aligned} &=x \times \frac{x}{x+1}\left(\frac{-1}{x^{2}}\right)+\log \left(1+\frac{1}{x}\right) \\\\ &=\frac{x^{2}}{x+1} \times \frac{-1}{x^{2}}+\log \left(1+\frac{1}{x}\right) \end{aligned}$$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\frac{-1}{x+1}+\log \left(1+\frac{1}{x}\right) \\\\ &\frac{d y}{d x}=y\left[\frac{-1}{x+1}+\log \left(1+\frac{1}{x}\right)\right] \end{aligned}$$=\left(1+\frac{1}{x}\right)^{x}\left[\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right]$Differentiation exercise Multiple choice question 7
Answer:$\frac{\log x}{(1+\log x)^{2}}$Hint:Differentiate the function w.r.t x
Given:$x^{y}=e^{x-y}$Solution: $x^{y}=e^{x-y}$Taking log on both sides
$\log x^{y}=\log \left(e^{x-y}\right)$$y \log x=(x-y) \log e \quad\left[\because \log m^{n}=n \log m\right]$$\begin{aligned} &y \log x=(x-y) \cdot 1\quad\quad\quad[\because \log e=1] \\\\ &y \log x=x-y \end{aligned}$$\begin{aligned} &x=y+y \log x \\\\ &x=y(1+\log x) \\\\ &y=\frac{x}{1+\log x} \end{aligned}$Differentiating y w.r.t x then
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x}{1+\log x}\right)$$=\frac{(1+\log x) \frac{d(x)}{d x}-x \frac{d}{d x}(1+\log x)}{(1+\log x)^{2}} \quad\left[\because \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}\right]$$=\frac{(1+\log x) \cdot 1-x\left\{\frac{d(1)}{d x}+\frac{d(\log x)}{d x}\right\}}{(1+\log x)^{2}}\left[\because \frac{d}{d x}\{f(x)+g(x)\}=\frac{d}{d x} f(x)+\frac{d}{d x} g(x), \frac{d(x)}{d x}=1\right]$$=\frac{1+\log x-x\left(0+\frac{1}{x}\right)}{(1+\log x)^{2}} \quad\left[\because \frac{d}{d x} \log x=\frac{1}{x}, \frac{d}{d x}(\operatorname{con} s \tan t)=0\right]$$=\frac{1+\log x-x \cdot \frac{1}{x}}{(1+\log x)^{2}}=\frac{1+\log x-1}{(1+\log x)^{2}}$$=\frac{\log x}{(1+\log x)^{2}}$Differentiation exercise Multiple choice question 9
Answer:$|\sec \theta|$Hint: Differentiate x and y w.r.t $\theta$, then divide and solve Given:$x=a \cos ^{3} \theta, y=a \sin ^{3} \theta, \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=$Solution: $\begin{aligned} &x=a \cos ^{3} \theta \\\\ &\frac{d x}{d \theta}=a \frac{d}{d \theta}\left(\cos ^{3} \theta\right) \\\\ &\frac{d x}{d \theta}=a(3) \cos ^{2} \frac{d}{d \theta}(\cos \theta) \end{aligned}$$\frac{d x}{d \theta}=-3 a \cos ^{2} \theta \sin \theta$ ....................(1)
$\begin{aligned} &y=a \sin \theta \\\\ &\frac{d y}{d \theta}=a \frac{d}{d \theta}\left(\sin ^{3} \theta\right) \end{aligned}$$=3 a \sin ^{2} \theta \frac{d}{d \theta}(\sin \theta)$$\frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta$ ......................(2)
Dividing (2) by (1) we get
$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 a \sin ^{2} \theta \cos \theta}{-3 a \cos ^{2} \theta \sin \theta}$$\begin{aligned} &\frac{d y}{d x}=\frac{\sin \theta}{-\cos \theta}=-\tan \theta \\\\ &\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=\sqrt{1+\tan ^{2} \theta}=\sqrt{\sec ^{2} \theta}=|\sec \theta| \end{aligned}$Differentiation exercise Multiple choice question 10
Answer:$\frac{-2}{1+x^{2}}$Hint:Differentiate y function w.r.t x
Given:$y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$Solution: Let
$y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$Differentiate y w.r.t x then
$\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right)$$=\frac{1}{\sqrt{1-\left(\frac{1-x^{2}}{1+x^{2}}\right)^{2}}}\left\{\frac{\left(1+x^{2}\right) \frac{d}{d x}\left(1-x^{2}\right)-\left(1-x^{2}\right) \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right\}$ $\left[\because \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}}, \frac{d}{d x}\left(\frac{u}{v}\right)=v \frac{d u}{d x}-u \frac{d v}{d x}\right]$$=\frac{1}{\sqrt{\frac{\left(1+x^{2}\right)^{2}-\left(1-x^{2}\right)^{2}}{\left(1+x^{2}\right)^{2}}}}\left[\frac{\left(1+x^{2}\right)\left\{\frac{d}{d x}(1)-\frac{d}{d x}\left(x^{2}\right)\right\}-\left(1-x^{2}\right)\left\{\frac{d}{d x}(1)+\frac{d}{d x}\left(x^{2}\right)\right\}}{\left(1+x^{2}\right)^{2}}\right]$ $\left[\because \frac{d}{d x}\{f(x)+g(x)\}=\frac{d}{d x} f(x)+\frac{d}{d x} g(x)\right]$$=\frac{\sqrt{\left(1+x^{2}\right)^{2}}}{\sqrt{\left(1+x^{4}+2 x^{2}\right)-\left(1+x^{4}-2 x^{2}\right)}}\left[\frac{\left(1+x^{2}\right)(0-2 x)-\left(1-x^{2}\right)(0+2 x)}{\left(1+x^{2}\right)^{2}}\right]$$\left[\frac{d}{d x}\left(x^{n}\right)=n x^{n-1},\left(a^{2}+b^{2}+2 a b\right)=(a+b)^{2} \&(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$$=\frac{\left(1+x^{2}\right)}{\sqrt{1+x^{4}+2 x^{2}-1-x^{4}+2 x^{2}}}\left[\frac{-2 x-2 x^{3}-\left(2 x-2 x^{3}\right)}{\left(1+x^{2}\right)^{2}}\right]$$=\frac{1+x^{2}}{\sqrt{4 x^{2}}}\left[\frac{-2 x-2 x^{3}-2 x+2 x^{3}}{\left(1+x^{2}\right)^{2}}\right]$$\begin{aligned} &\left.=\frac{1+x^{2}}{\sqrt{(2 x)^{2}}\left[\frac{-4 x}{\left(1+x^{2}\right)^{2}}\right.}\right]=\frac{\left(1+x^{2}\right)}{2 x} \times \frac{-4 x}{\left(1+x^{2}\right)^{2}}=\frac{-2}{\left(1+x^{2}\right)} \\\\ &\therefore \frac{d y}{d x}=\frac{-2}{1+x^{2}} \end{aligned}$Differentiation exercise Multiple choice question 11
Answer:Does not exist
Hint:$\sec ^{-1} \alpha$is not defined
Given:$\sec ^{-1}\left(\frac{1}{2 x^{2}+1}\right) \text { W.r.t } \sqrt{1+3 x} \text { at } x=-\frac{1}{3}$Solution: We know that
$\sec ^{-1} \alpha$ is not defined for
$\alpha \in(-1,1)$Here for
$x=-\frac{1}{3}, \frac{1}{2 x^{2}+1}=\frac{9}{11} \in(-1,1)$$\therefore \sec ^{-1}\left(\frac{1}{2 x^{2}+1}\right)$ Is not defined at
$x=-\frac{1}{3}$Derivative of
$\sec ^{-1}\left(\frac{1}{2 x^{2}+1}\right)$ does not exist at
$x=-\frac{1}{3}$Differentiation exercise Multiple choice question 12
Answer: $-1$Hint: Differentiate the function w.r.t x
Given:$\sqrt{x}+\sqrt{y}=1, \frac{d y}{d x} a t\left(\frac{1}{4}, \frac{1}{4}\right)$Solution: $\sqrt{x}+\sqrt{y}=1$Differentiating w.r.t x
$\begin{aligned} &\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0 \\\\ &\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=\frac{-1}{2 \sqrt{x}} \end{aligned}$$\frac{d y}{d x}=-\frac{1}{2 \sqrt{x}} \times \frac{2 \sqrt{y}}{1}=\frac{-\sqrt{y}}{\sqrt{x}}$Now,
$\left[\frac{d y}{d x}\right]_{\left(\frac{1}{4} \cdot \frac{1}{4}\right)}=\frac{-\sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}}=-1$Differentiation exercise Multiple choice question 13
Answer:$-1$Hint: Differentiate the function w.r.t x
Given:$\sin (x+y)=\log (x+y)$Solution: $\begin{aligned} &\sin (x+y)=\log (x+y) \\\\ &\cos (x+y)\left(1+\frac{d y}{d x}\right)=\frac{1}{(x+y)}\left(1+\frac{d y}{d x}\right) \end{aligned}$$\begin{aligned} &\cos (x+y)+\cos (x+y) \frac{d y}{d x}=\frac{1}{(x+y)}+\frac{1}{(x+y)} \frac{d y}{d x} \\\\ &\cos (x+y) \frac{d y}{d x}-\frac{1}{(x+y)} \frac{d y}{d x}=\frac{1}{(x+y)}-\cos (x+y) \end{aligned}$$\begin{aligned} &{\left[\cos (x+y)-\frac{1}{(x+y)}\right] \frac{d y}{d x}=\frac{1}{(x+y)}-\cos (x+y)} \\\\ &-\left[\frac{1}{(x+y)}-\cos (x+y)\right] \frac{d y}{d x}=\frac{1}{(x+y)}-\cos (x+y) \\\\ &\frac{d y}{d x}=-1 \end{aligned}$Differentiation exercise Multiple choice question 14
Answer: 1
Hint: Differentiate the function w.r.t x
Given:$u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), v=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$Solution: $u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), v=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$Differentiating u w.r.t x then
$\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\right)$$=\frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}}}\left[\frac{\left(1+x^{2}\right) \frac{d}{d x}(2 x)-2 x \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right]$ $\left[\begin{array}{l} \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\\\ \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}} \end{array}\right]$$=\frac{1}{\sqrt{\frac{\left(1+x^{2}\right)^{2}-(2 x)^{2}}{\left(1+x^{2}\right)^{2}}}}\left[\frac{\left(1+x^{2}\right) 2 \cdot \frac{d(x)}{d x}-2 x\left\{\frac{d}{d x}(1)+\frac{d}{d x}\left(x^{2}\right)\right\}}{\left(1+x^{2}\right)^{2}}\right]$ $\left[\begin{array}{l} \because \frac{d}{d x}\{f(x)+g(x)\}=\frac{d}{d x} f(x)+\frac{d}{d x} g(x) \\\\ \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right) \end{array}\right]$$=\frac{\sqrt{\left(1+x^{2}\right)^{2}}}{\sqrt{1+x^{4}+2 x^{2}-4 x^{2}}} \times\left[\frac{\left(1+x^{2}\right) 2 \cdot 1-2 x(0+2 x)}{\left(1+x^{2}\right)^{2}}\right]$ $\left[\because \frac{d}{d x}(\operatorname{cons} \tan t)=0, \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]$$=\frac{\left(1+x^{2}\right)}{\sqrt{1+x^{2}-2 x^{2}}}\left[\frac{\left(1+x^{2}\right) 2-2 x(2 x)}{\left(1+x^{2}\right)^{2}}\right]$$=\frac{\left(1+x^{2}\right)}{\sqrt{\left(1-x^{2}\right)^{2}}}\left[\frac{2+2 x^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}\right] \quad\quad\quad\left[\because a^{2}+b^{2}-2 a b=(a-b)^{2}\right]$$\begin{aligned} &=\frac{\left(1+x^{2}\right)}{\left(1-x^{2}\right)} \times \frac{2-2 x^{2}}{\left(1+x^{2}\right)^{2}}=\frac{2\left(1-x^{2}\right)}{\left(1-x^{2}\right)\left(1+x^{2}\right)}=\frac{2}{1+x^{2}}\\\\ &\therefore \frac{d u}{d x}=\frac{2}{1+x^{2}} &\text { } \end{aligned}$................(1)
Differentiating v w.r.t x then,
$\frac{d v}{d x}=\frac{d}{d x}\left(\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right)$$=\frac{1}{1+\left(\frac{2 x}{1-x^{2}}\right)^{2}}\left[\frac{\left(1-x^{2}\right) \frac{d}{d x}(2 x)-\frac{d}{d x}\left(1-x^{2}\right) \cdot 2 x}{\left(1-x^{2}\right)^{2}}\right]$ $\left[\because \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}, \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}\right]$$=\frac{1}{\frac{\left(1-x^{2}\right)^{2}+(2 x)^{2}}{\left(1-x^{2}\right)^{2}}}\left[\frac{\left(1-x^{2}\right) 2 \frac{d}{d x}(x)-2 x\left\{\frac{d}{d x}(1)-\frac{d}{d x}\left(x^{2}\right)\right\}}{\left(1-x^{2}\right)^{2}}\right]$ $\left[\begin{array}{l} \frac{d}{d x}\{f(x)+g(x)\}=\frac{d}{d x} f(x)+\frac{d}{d x} g(x) \\\\ \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right) \end{array}\right]$$=\frac{\left(1-x^{2}\right)^{2}}{1+x^{4}-2 x^{2}+4 x^{2}}\left[\frac{\left(1-x^{2}\right) \cdot 2 \cdot 1-2 x(0-2 x)}{\left(1-x^{2}\right)^{2}}\right]$ $\left[\begin{array}{l} \because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}, \frac{d}{d x}(\operatorname{cons} \tan t)=0, \frac{d(x)}{d x}=1 \\\\ (a-b)^{2}=a^{2}+b^{2}-2 a b \end{array}\right]$$=\frac{1}{1+x^{4}+2 x^{2}}\left[\frac{2-2 x^{2}+4 x^{2}}{1}\right]$$=\frac{2+2 x^{2}}{\left(1+x^{2}\right)^{2}}=\frac{2\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}} \quad\quad\quad\left[\because a^{2}+b^{2}+2 a b=(a+b)^{2}\right]$$\therefore \frac{d v}{d x}=\frac{2}{1+x^{2}}$ ..............................(2)
Thus
$\frac{d y}{d x}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{d u}{d x} \times \frac{d x}{d v}$$=\frac{2}{1+x^{2}} \times \frac{1+x^{2}}{2}$ [Using (1) and (2)]
$\therefore \frac{d y}{d x}=1$Differentiation exercise Multiple choice question 15
Answer:$-\frac{1}{2}$Hint: Differentiate the function w.r.t x
Given:$\frac{d}{d x}\left[\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\right]$Solution: $u=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$$=\tan ^{-1}\left(\frac{\cos ^{2}\left(\frac{x}{2}\right)-\sin ^{2}\left(\frac{x}{2}\right)}{\cos ^{2}\left(\frac{x}{2}\right)-\sin ^{2}\left(\frac{x}{2}\right)+2 \sin \frac{x}{2} \cos \frac{x}{2}}\right)$$=\tan ^{-1} \frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}$$=\tan ^{-1}\left[\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right]$$u=\tan ^{-1}\left[\frac{\tan \frac{\pi}{4}-\tan \frac{x}{2}}{1+\tan \frac{\pi}{4} \times \tan \frac{x}{2}}\right]$$=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right]$$\begin{aligned} &u=\frac{\pi}{4}-\frac{x}{2} \\\\ &\frac{d u}{d x}=0-\frac{1}{2} \\\\ &\frac{d u}{d x}=-\frac{1}{2} \end{aligned}$Differentiation exercise Multiple choice question 16
Answer: $\frac{x^{2}-1}{x^{2}-4}$Hint:Differentiate the function w.r.t x
Given:$\frac{d}{d x}\left[\log \left\{e^{x}\left(\frac{x-2}{x+2}\right)^{\frac{3}{4}}\right\}\right]$Solution: $y=\frac{d}{d x}\left[\log \left(e^{x}\left(\frac{x-2}{x+2}\right)^{2}\right)\right]$$=\frac{d}{d x}\left[x \log e+\frac{3}{4} \log \left(\frac{x-2}{x+2}\right)\right]$$y=\frac{d}{d x}\left[x+\frac{3}{4} \log \left(\frac{x-2}{x+2}\right)\right]$$\frac{d y}{d x}=1+\frac{3}{4\left(\frac{x-2}{x+2}\right)} \times \frac{(x+2)-(x-2)}{(x+2)^{2}}$$\begin{aligned} &=1+\frac{3}{4} \frac{(x+2)}{(x-2)} \times \frac{x+2-x+2}{(x+2)^{2}} \\\\ &=1+\frac{3}{4} \frac{(x+2)}{(x-2)} \times \frac{4}{(x+2)} \end{aligned}$$\begin{aligned} &=1+\frac{3}{\left(x^{2}-4\right)} \\\\ &\frac{d y}{d x}=\frac{x^{2}-4+3}{x^{2}-4}=\frac{x^{2}-1}{x^{2}-4} \end{aligned}$Differentiation exercise Multiple choice question 17
Answer:$\frac{\cos 2 x}{2 y-1}$Hint: Differentiate the function w.r.t x
Given:$y=\sqrt{\sin x+y}$Solution: $y=\sqrt{\sin x+y}$Squaring both side
$\begin{aligned} &y^{2}=\sin x+y \\\\ &y^{2}-y=\sin x \\\\ &2 y \frac{d y}{d x}-\frac{d y}{d x}=\cos x \end{aligned}$ Differentiation exercise Multiple choice question 18
Answer:$-\frac{y}{x}$Hint:Differentiate the function w.r.t x
Given:$3 \sin (x y)+4 \cos (x y)=5$Solution: $3 \sin (x y)+4 \cos (x y)=5$$\begin{aligned} &3 \cos (x y)\left[x \frac{d y}{d x}+y\right]-4 \sin (x y)\left[x \frac{d y}{d x}+y\right]=0 \\\\ &{\left[x \frac{d y}{d x}+y\right][3 \cos (x y)-4 \sin (x y)]=0} \end{aligned}$$\begin{aligned} &x \frac{d y}{d x}+y=0 \\\\ &x \frac{d y}{d x}=-y \\\\ &\frac{d y}{d x}=-\frac{y}{x} \end{aligned}$Differentiation exercise Multiple choice question 19
Answer:$\frac{\sin ^{2}(a+y)}{\sin a}$Hint: Differentiate the function w.r.t x
Given:$\sin y=x \sin (a+y), \text { then } \frac{d y}{\alpha x}$Solution: $\begin{aligned} &\sin y=x \sin (a+y) \\\\ &\frac{d}{d x}(\sin y)=\frac{d}{d x}[(a+y) \sin x] \end{aligned}$$\cos y \frac{d y}{d x}=\sin (a+y) \frac{d}{d x}(x)+x \frac{d}{d x}[\sin (a+y)]$$\begin{aligned} &=\sin (a+y) \times 1+x \cos (a+y) \frac{d y}{d x} \\ &=\sin (a+y)+x \cos (a+y) \frac{d y}{d x} \end{aligned}$$\cos y \frac{d y}{d x}=-x \cos (a+y) \frac{d y}{d x}=\sin (a+y)$$\left(\cos y-\frac{\sin y}{\sin (a+y)} \times \cos (a+y)\right) \frac{d y}{d x}=\sin (a+y)$ $\left[\begin{array}{l} \sin y=2 \sin x \cos x \\\\ x=\frac{\sin y}{\sin (a+y)} \end{array}\right]$$\left(\frac{\sin (a+y) \cos y-\sin y \cos (a+y)}{\sin (a+y)}\right) \frac{d y}{d x}=\sin (a+y)$$\begin{aligned} &\frac{\sin (a+y-y)}{\sin (a+y)} \times \frac{d y}{d x}=\sin (a+y) \\\\ &\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a} \end{aligned}$Differentiation exercise Multiple choice question 20
Answer: 2
Hint: Differentiate the function w.r.t $\cos ^{-1}x$ Given: $\cos ^{-1}(2x^{2}-1)$ With respect to
$\cos ^{-1}x$ Solution: $\begin{aligned} &u=\cos ^{-1}\left(2 x^{2}-1\right) \\\\ &x=\cos \theta \\\\ &\theta=\cos ^{-1} x \\\\ &\frac{d \theta}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}$$\begin{aligned} &u=\cos ^{-1}(\cos 2 \theta) \\\\ &u=2 \theta \\\\ &\frac{d u}{d x}=2 \frac{d \theta}{d x} \\\\ &\frac{d u}{d x}=\frac{-2}{\sqrt{1-x^{2}}} \end{aligned}$ ...............(1)
$\begin{aligned} &v=\cos ^{-1} x \\\\ &v=\cos ^{-1}(\cos \theta), v=\theta \\\\ &\frac{d v}{d x}=\frac{d \theta}{d x} \\\\ &\frac{d v}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}$ ..................(2)
Divide (1) by (2)
$\begin{aligned} &\frac{d u}{d x}=\frac{-2}{\frac{d v}{d x}}{\sqrt{1-x^{2}}} \times \frac{\sqrt{1-x^{2}}}{-1} \\\\ &\frac{d u}{d v}=2 \end{aligned}$Differentiation exercise Multiple choice question 21
Answer:$-1 \text { for } x<-3$Hint:Differentiate the function w.r.t x
Given:$f(x)=\sqrt{x^{2}+6 x+9}, f^{\prime}(x)$Solution: $\begin{aligned} &f(x)=\sqrt{x^{2}+6 x+9} \\\\ &x^{2}+6 x+9=(x+3)(x+3)=(x+3)^{2} \end{aligned}$$\begin{aligned} &f(x)=\sqrt{(x+3)^{2}}=|x+3| \\\\ &f(x)=\left\{\begin{array}{cc} x+3 & x \geq-3 \\\\ -x-3 & x<-3 \end{array}\right. \end{aligned}$$\begin{aligned} &f^{\prime}(x)=\left\{\begin{array}{cl} 1 & x \geq-3 \\\\ -1 & x<-3 \end{array}\right. \\\\ &f^{\prime}(x)=-1 \text { For } x<-3 \end{aligned}$Differentiation exercise Multiple choice question 22
Answer:$-2 x+9, \text { If } 4<x<5$Hint: Differentiate the function w.r.t x
Given:$f(x)=\left|x^{2}-9 x+20\right| \text { Then } f^{\prime}(x)=$Solution: $f(x)=\left|x^{2}-9 x+20\right|$$f(x)=\left\{\begin{array}{cc} x^{2}-9 x+20 & -\infty<x \leq 4 \\\\ -\left(x^{2}-9 x+20\right) & 4<x<5 \\\\ x^{2}-9 x+20 & 5 \leq x<\infty \end{array}\right.$$f^{\prime}(x)= \begin{cases}2 x-9 & \infty<x \leq 4 \\\\ -2 x+9 & 4<x<5 \\\\ 2 x-9 & 5 \leq x<\infty\end{cases}$$f^{\prime}(x)=-2 x+9 \text { for } 4<x<5$Differentiation exercise Multiple choice question 23
Answer: None of these
Hint: In case of function of one variable it’s a function that doesn’t have finite derivation
Given:$f(x)=\sqrt{x^{2}-10 x+25}, f^{\prime}(x)=$Solution: $f(x)=\sqrt{x^{2}-10 x+25}$$\begin{aligned} &=\sqrt{(x-5)^{2}} \\\\ &=|x-5| \end{aligned}$$f(x)=\left\{\begin{array}{cll} x-5 & \text { for } & x>5 \\\\ -(x-5) & \text { for } & x<5 \end{array}\right.$$L H D=\lim _{x \rightarrow 5^{-}} \frac{f(x)-f(a)}{x-a}$$=\lim _{x \rightarrow 5^{-}} \frac{\sqrt{x^{2}-10 x+25}-\sqrt{5^{2}-10(5)+25}}{x-5}$$=\lim _{x \rightarrow 5^{-}} \frac{|x-5|}{(x-5)}=\lim _{x \rightarrow 5^{-}} \frac{-(x-5)}{x-5}=-1$$R H D=\lim _{x \rightarrow 5^{+}} \frac{f(x)-f(a)}{x-a}$$=\lim _{x \rightarrow 5^{+}} \frac{\sqrt{x^{2}-10 x+25}-\sqrt{5^{2}-10(5)+25}}{x-5}$$\begin{aligned} &=\lim _{x \rightarrow 5^{+}} \frac{|x-5|}{x-5} \\\\ &=\lim _{x \rightarrow 5^{+}} \frac{x-5}{x-5} \\\\ &=1 \end{aligned}$$L H D \neq R H D,$ so function is not differentiable .
Differentiation exercise Multiple choice question 25
Answer: 0
Hint: Differentiate the function w.r.t x
Given:$f(x)=\left(\frac{x^{l}}{x^{m}}\right)^{l+m}\left(\frac{x^{m}}{x^{n}}\right)^{m+n}\left(\frac{x^{n}}{x^{3}}\right)^{n+l}$Solution: $f(x)=\left(\frac{x^{l}}{x^{m}}\right)^{l+m}\left(\frac{x^{m}}{x^{n}}\right)^{m+n}\left(\frac{x^{n}}{x^{l}}\right)^{n+l}$$\begin{aligned} &=x^{(l-m)(l+m)} \times x^{(m-n)(m+n)} \times x^{(n-l)(n+l)} \\\\ &=x^{l^{2}-m^{2}} \times x^{m^{2}-n^{2}} \times x^{n^{2}-l^{2}} \end{aligned}$$\begin{aligned} &f(x)=x^{\left(l^{2}-m^{2}+m^{2}-n^{2}+n^{2}-l^{2}\right)}=x^{0} \\\\ &f(x)=1 \\\\ &f^{\prime}(x)=0 \end{aligned}$Differentiation exercise Multiple choice question 26
Answer: 0
Hint: Differentiate the function w.r.t x
Given:$y=\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}+\frac{1}{1+x^{b-a}+x^{c-a}}$Solution: $y=\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}+\frac{1}{1+x^{b-a}+x^{c-a}}$$=\frac{1}{1+\frac{x^{a}}{x^{b}}+\frac{x^{c}}{x^{b}}}+\frac{1}{1+\frac{x^{b}}{x^{c}}+\frac{x^{a}}{x^{c}}}+\frac{1}{1+\frac{x^{b}}{x^{a}}+\frac{x^{c}}{x^{a}}}$$=\frac{x^{b}}{x^{a}+x^{b}+x^{c}}+\frac{x^{c}}{x^{a}+x^{b}+x^{c}}+\frac{x^{a}}{x^{a}+x^{b}+x^{c}}$$\begin{aligned} &=\frac{x^{b}+x^{c}+x^{a}}{x^{a}+x^{b}+x^{c}} \\\\ &y=1 \\\\ &\frac{d y}{d x}=\frac{d(1)}{d x}=0 \end{aligned}$Differentiation exercise Multiple choice question 27
Answer:$\frac{x^{2}}{y^{2}} \sqrt{\frac{1-y^{6}}{1-x^{6}}}$Hint: Differentiate the function w.r.t x
Given:$\sqrt{1-x^{6}}+\sqrt{1-y^{6}}=a^{3}\left(x^{3}-y^{3}\right)$Solution: We have,
$\sqrt{1-x^{6}}+\sqrt{1-y^{6}}=a^{3}\left(x^{3}-y^{3}\right)$Putting
$x^{3}=\sin A, y^{3}=\sin B$$\begin{aligned} &\sqrt{1-\sin ^{2} A}+\sqrt{1-\sin ^{2} B}=a(\sin A-\sin B) \\\\ &\cos A+\cos B=a(\sin A-\sin B) \end{aligned}$$\begin{aligned} &2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)=2 a \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right) \\\\ &\cot \left(\frac{A-B}{2}\right)=a^{3} \end{aligned}$$\begin{aligned} &\frac{A-B}{2}=\cot ^{-1}\left(a^{3}\right) \\\\ &A-B=2 \cot ^{-1}\left(a^{3}\right) \\\\ &\sin ^{-1} x^{3}-\sin ^{-1} y^{3}=2 \cot ^{-1}\left(a^{3}\right) \end{aligned}$$\begin{aligned} &\frac{1}{\sqrt{1-x^{6}}} \times \frac{d}{d x}\left(x^{3}\right)-\frac{1}{\sqrt{1-y^{6}}} \times \frac{d}{d x}\left(y^{3}\right)=0 \\\\ &\frac{1}{\sqrt{1-x^{6}}} \times 3 x^{2}-\frac{1}{\sqrt{1-y^{6}}} \times 3 y^{2} \times \frac{d y}{d x}=0 \end{aligned}$$\frac{d y}{d x}=\frac{x^{2}}{y^{2}} \sqrt{\frac{1-y^{6}}{1-x^{6}}}$Differentiation exercise Multiple choice question 28
Answer: 1
Hint: Differentiate the function w.r.t x
Given:$y=\log \sqrt{\tan x}$Solution: $\begin{aligned} &y=\log \sqrt{\tan x} \\\\ &\frac{d y}{d x}=\frac{1}{\sqrt{\tan x}} \times \frac{d}{d x}(\sqrt{\tan x}) \end{aligned}$$=\frac{1}{\sqrt{\tan x}} \times \frac{1}{2 \sqrt{\tan x}} \times \frac{d}{d x}(\tan x)$$\frac{d y}{d x}=\frac{\sec ^{2} x}{2 \tan x}$Now
$\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{4}}=\frac{\left(\sec \frac{\pi}{4}\right)^{2}}{2 \tan \left(\frac{\pi}{4}\right)}=\frac{2}{2 \times 1}=1$Differentiation exercise Multiple choice question 29
Answer:$\frac{y}{x}$Hint: Differentiate the function w.r.t x
Given:$\sin ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=\log a$Solution: $\sin ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=\log a$$\frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\sin \log a$$\frac{\left(x^{2}+y^{2}\right)\left(2 x-2 y \frac{d y}{d x}\right)-\left(x^{2}-y^{2}\right)\left(2 x+2 y \frac{d y}{d x}\right)}{\left(x^{2}+y^{2}\right)^{2}}=0$$\frac{2 x^{3}-2 x^{2} y d y+2 x y^{2}-2 y^{3} \frac{d y}{d x}-2 x^{3}-2 x^{2} y \frac{d y}{d x}+2 x y^{2}+2 y^{3} \frac{d y}{d x}}{\left(x^{2}+y^{2}\right)^{2}}=0$$\begin{aligned} &-4 x^{2} y \frac{d y}{d x}+4 x y^{2}=0 \\\\ &-4 x^{2} y \frac{d y}{d x}=-4 x y^{2} \\\\ &\frac{d y}{d x}=\frac{4 x y^{2}}{4 x^{2} y}=\frac{y}{x} \end{aligned}$Differentiation exercise Multiple choice question 30
Answer:$\frac{\cos ^{2}(a+y)}{\cos a}$Hint:Differentiate the function w.r.t x
Given:$\sin y=x \cos (a+y)$Solution: $\begin{aligned} &\sin y=x \cos (a+y) \\\\ &\frac{d}{d x}(\sin y)=\frac{d}{d x}[x \cos (a+y)] \end{aligned}$$\begin{aligned} &\cos y \frac{d y}{d x}=1 \times \cos (a+y)-x \sin (a+y) \frac{d}{d x}(a+y) \\\\ &\cos y \frac{d y}{d x}=\cos (a+y)-x \sin (a+y) \frac{d y}{d x} \end{aligned}$$\begin{aligned} &\cos y \frac{d y}{d x}+x \sin (a+y) \frac{d y}{d x}=\cos (a+y) \\\\ &{[\cos y+x \sin (a+y)] \frac{d y}{d x}=\cos (a+y)} \end{aligned}$$\left[\cos y+\frac{\sin y}{\cos (a+y)} \times \sin (a+y)\right] \frac{d y}{d x}=\cos (a+y)$ $\left[\begin{array}{l} \sin y=x \cos (a+y) \\\\ x=\frac{\sin y}{\cos (a+y)} \end{array}\right]$$\left[\frac{\cos (a+y) \cos y+\sin y \sin (a+y)}{\cos (a+y)}\right] \frac{d y}{d x}=\cos (a+y)$$\begin{aligned} &\frac{\cos (a+y-y)}{\cos (a+y)} \times \frac{d y}{d x}=\cos (a+y) \\\\ &\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\cos a} \end{aligned}$Differentiation exercise Multiple choice question 31
Answer:$-\frac{4 x}{1-x^{4}}$Hint:Differentiate the function w.r.t x
Given:$y=\log \left(\frac{1-x^{2}}{1+x^{2}}\right)$Solution: $y=\log \left(\frac{1-x^{2}}{1+x^{2}}\right)$$\frac{d y}{d x}=\frac{1}{\frac{1-x^{2}}{1+x^{2}}} \frac{d}{d x}\left(\frac{1-x^{2}}{1+x^{2}}\right)$$=\frac{1+x^{2}}{1-x^{2}}\left[\frac{\left(1+x^{2}\right)(-2 x)-\left(1-x^{2}\right)(2 x)}{\left(1+x^{2}\right)^{2}}\right]$$\begin{aligned} &=\frac{1}{1-x^{2}}\left[\frac{-2 x-2 x^{3}-2 x+2 x^{3}}{1+x^{4}}\right] \\\\ &\frac{d y}{d x}=\frac{-4 x}{1-x^{4}} \end{aligned}$Differentiation exercise Multiple choice question 32
Answer: 1
Hint: Differentiate the function w.r.t x
Given:$y=\tan ^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)$Solution: $y=\tan ^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)$$\frac{d y}{d x}=\frac{1}{1+\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)^{2}} \frac{d}{d x}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)$$\frac{d y}{d x}=\frac{(\cos x-\sin x)^{2}}{(\cos x-\sin x)^{2}+(\sin x+\cos x)^{2}} \times$ $\left[\frac{(\cos x-\sin x) \frac{d}{d x}(\sin x+\cos x)-(\sin x+\cos x) \frac{d}{d x}(\cos x-\sin x)}{(\cos x-\sin x)^{2}}\right]$$=\frac{(\cos x-\sin x)^{2}}{(\cos x-\sin x)^{2}+(\sin x+\cos x)^{2}}\left[\frac{(\cos x-\sin x)(\cos x-\sin x)-(\sin x+\cos x)(-\sin x-\cos x)}{(\cos x-\sin x)^{2}}\right]$$=\frac{(\cos x-\sin x)^{2}}{(\cos x-\sin x)^{2}+(\sin x+\cos x)^{2}}\left[\frac{(\cos x-\sin x)(\cos x-\sin x)+(\sin x+\cos x)(\sin x+\cos x)}{(\cos x-\sin x)^{2}}\right]$$\begin{aligned} &=\frac{(\cos x-\sin x)^{2}}{(\cos x-\sin x)^{2}(\sin x+\cos x)^{2}} \times \frac{(\cos x-\sin x)^{2}+(\sin x+\cos x)^{2}}{(\cos x-\sin x)^{2}} \\ &\frac{d y}{d x}=1 \end{aligned}$Differentiation exercise Multiple choice question 33
Answer:$\text { (c) } \frac{d y}{d x}=\frac{y-1}{x+1}$Hint: Differentiate the function w.r.t x
Given:$\sec ^{-1}\left(\frac{1+x}{1-y}\right)$Solution: $\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a$$\begin{aligned} &\sec a=\frac{1+x}{1-y} \\\\ &(1-y) \sec a=1+x \end{aligned}$Differentiate w.r.t x
$-\sec a \frac{d y}{d x}=1 \quad\quad\quad\quad\left[\frac{d(1)}{d x}=0, \frac{d(x)}{d x}=1\right]$$\frac{d y}{d x}=\frac{-1}{\sec a}$$\begin{aligned} &=\frac{-(1-y)}{(1+x)} \\\\ &=\frac{y-1}{1+x} \end{aligned}$Hence, option (c) is correct.
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