Choose the correct alternative in the following:
Question: Choose the correct alternative in the following: For the curve $\cdot \sqrt{x}+\sqrt{y}=1, \frac{d y}{d x} \cdot$ at $(1 / 4,1 / 4)$ is A. $1 / 2$ B. 1 C. $-1$ D. 2 Solution: $\sqrt{x}+\sqrt{y}=1$ Differentiating w.r.t $\mathrm{x}$ we get, $\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}(\sqrt{\mathrm{x}})+\frac{\mathrm{d}}{\mathrm{dx}}(\sqrt{\mathrm{y}})=\frac{\mathrm{d}}{\mathrm{dx}}(1)$ $\Rightarrow \frac{1}{2 \sqrt{\mathrm{x}}}+\frac{1}{2 \sqrt{\mathrm{y}}} \cdot \frac{\mathrm{dy}}{\mat...
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Question: Choose the correct alternative in the following The derivative of $\sec ^{-1}\left(\frac{1}{2 x^{2}+1}\right)$ with respect to $\sqrt{1+3 x}$ at $x=-1 / 3$ A. does not exist B. 0 C. $1 / 2$ D. $1 / 3$ Solution: Let $u=\sec ^{-1}\left(\frac{1}{2 x^{2}+1}\right)$ and $v=\sqrt{1+3 x}$ $\left(\frac{d u}{d v}\right)_{x=-\frac{1}{3}}=?$ Considering $u$, $u=\sec ^{-1}\left(\frac{1}{2 x^{2}+1}\right)$ Put $x=\cos \theta$ $\theta=\cos ^{-1} x \cdots(1)$ $u=\sec ^{-1}\left(\frac{1}{2 \cos ^{2} \...
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Question: Choose the correct alternative in the following: If A. $-\frac{2}{1+\mathrm{x}^{2}}$ B. $\frac{2}{1+x^{2}}$ C. $\frac{1}{2-x^{2}}$ D. $\frac{2}{2-x^{2}}$ Solution: $y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$ Put $x=\tan \theta \Rightarrow \theta=\tan ^{-1} x$ $y=\sin ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)$ $y=\sin ^{-1}(\cos 2 \theta) \because \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\cos 2 \theta$ $y=\sin ^{-1}\left(\sin \left(\frac{\pi}{2}-2 \thet...
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Question: Choose the correct alternative in the following: If $x=a \cos ^{3} \theta, y=a \sin ^{3} \theta$, then $\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=$ A. $\tan ^{2} \theta$ B. $\sec ^{2} \theta$ C. $\sec \theta$ D. $|\sec \theta|$ Solution: We are given that $\mathrm{x}=\mathrm{a} \cdot \cos ^{3} \theta, \mathrm{y}=\mathrm{a} \cdot \sin ^{3} \theta$ $\sqrt{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}}=?$ Now, we know $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{...
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Question: Choose the correct alternative in the following: Given $f(x)=4 x^{8}$, then A. $f^{\prime}\left(\frac{1}{2}\right)=f^{\prime}\left(-\frac{1}{2}\right)$ B. $f\left(\frac{1}{2}\right)=f^{\prime}\left(-\frac{1}{2}\right)$ C. $f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{2}\right)$ D. $f\left(\frac{1}{2}\right)=f^{\prime}\left(-\frac{1}{2}\right)$ Solution: $f(x)=4 x^{8}$ $f^{\prime}(x)=32 x^{7}$ Consider option (A) $\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=32\left(\frac{1}{2}\right)^{7}...
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Question: Choose the correct alternative in the following: If $x^{y}=e^{x-y}$, then $\frac{d y}{d x}$ is A. $\frac{1+\mathrm{x}}{1+\log \mathrm{x}}$ B. $\frac{1-\log x}{1+\log x}$ C. not defined D. $\frac{\log x}{(1+\log x)^{2}}$ Solution: $x^{y}=e^{x-y}$ Taking log both sides we get $\log x^{y}=\log e^{x-y}$ $y \log x=(x-y) \log e$ $y \log x=(x-y) \because \log e=1$ $y=\frac{x}{\log x+1}$ Differentiating w.r.t $x$ we get, $\frac{d y}{d x}=\frac{1 \cdot(\log x+1)-\frac{1}{x} \cdot x}{(\log x+1)^...
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Question: Choose the correct alternative in the following: If $y=\left(1+\frac{1}{x}\right)^{x}$, then $\frac{d y}{d x}=$ A. $\left(1+\frac{1}{x}\right)^{x}\left\{\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right\}$ B. $\left(1+\frac{1}{x}\right)^{x} \log \left(1+\frac{1}{x}\right)$ C. $\left(1+\frac{1}{x}\right)^{x}\left\{\log (x+1)-\frac{x}{x+1}\right\}$ D. $\left(1+\frac{1}{x}\right)^{x}\left\{\log \left(x+\frac{1}{x}\right)+\frac{1}{x+1}\right\}$ Solution: Given $y=\left(1+\frac{1}{x}\righ...
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Question: Choose the correct alternative in the following: If $f(x)=\tan ^{-1} \sqrt{\frac{1+\sin x}{1-\sin x}}, 0 \leq x \leq \pi / 2$, then $f^{\prime}(\pi / 6)$ is A. $-1 / 4$ B. $-1 / 2$ C. $1 / 4$ D. $1 / 2$ Solution: $f(x)=\tan ^{-1} \sqrt{\frac{1+\sin x}{1-\sin x}}$ $=\tan ^{-1} \sqrt{\frac{1+2 \cdot \sin \frac{x}{2} \cos \frac{x}{2}}{1-2 \cdot \sin \frac{x}{2} \cos \frac{x}{2}}}$ $\because \sin 2 x=2 \sin x \cos x$ $\Rightarrow \sin x=2 \sin x / 2 \cos x / 2$ $=\tan ^{-1} \sqrt{\frac{\si...
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Question: Choose the correct alternative in the following: Differential coefficient of $\sec \left(\tan ^{-1} x\right)$ is A. $\frac{\mathrm{x}}{1+\mathrm{x}^{2}}$ B. $x \sqrt{1+x^{2}}$ C. $\frac{1}{\sqrt{1+x^{2}}}$ D. $\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}$ Solution: Let $f(x)=\sec \left(\tan ^{-1} x\right)$ Let $\theta=\tan ^{-1} x$ $\frac{d \theta}{d x}=\frac{1}{1+x^{2}}$ .....(1) Now $\theta=\tan ^{-1} x$ $=x=\tan \theta$ $=\sqrt{1+x^{2}}=\sec \theta \because \sec ^{2} \theta-\tan ^{2} ...
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Question: Choose the correct alternative in the following: The derivative of the function $\cos ^{-1}\left\{(\cos 2 x)^{1 / 2}\right\}$ at $x=\pi / 6$ is A. $(2 / 3)^{1 / 2}$ B. $(1 / 3)^{1 / 2}$ C. $3^{1 / 2}$ D. $6^{1 / 2}$ Solution: $\mathrm{f}^{\prime}(\mathrm{x})=-\frac{1}{\sqrt{1-\left[(\cos 2 \mathrm{x})^{\frac{1}{2}}\right]^{2}}} \cdot \frac{1}{2 \sqrt{\cos 2 \mathrm{x}}} \cdot(-\sin 2 \mathrm{x}) \cdot 2$ $=\frac{1}{\sqrt{1-\cos 2 x}} \cdot \frac{1}{\sqrt{\cos 2 x}} \cdot(\sin 2 x)$ Put...
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Question: Choose the correct alternative in the following: The differential coefficient of $f(\log x)$ with respect to $x$, where $f(x)=\log x$ is A. $\frac{\mathrm{x}}{\log \mathrm{x}}$ B. $\frac{\log x}{x}$ C. $(x \log x)^{-1}$ D. none of these Solution: Given: $f(x)=\log x$ $\therefore f(\log x)=\log (\log x)$ $f^{\prime}(\log x)=\frac{d}{d x} \log (\log x)$ $f^{\prime}(\log x)=\frac{1}{\log x} \cdot \frac{1}{x}=\frac{1}{x \log x}$ $\therefore f^{\prime}(\log x)=(x \log x)^{-1}$...
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Question: Choose the correct alternative in the following: If $f(x)=\log _{x} 2(\log x)$, then $f^{\prime}(x)$ at $x=e$ is A. 0 B. 1 C. $1 / \mathrm{e}$ D $1 / 2 \mathrm{e}$ Solution: $f(x)=\log _{x} 2(\log x)$ Changing the base, we get $\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\log (\log \mathrm{x})}{\log \mathrm{x}^{2}}$ $\because \log _{b} a=\frac{\log a}{\log b}$ $\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\log (\log \mathrm{x})}{2 \cdot \log \mathrm{x}}$ So, $f^{\prime}(x)=\frac{1}{2}\left\{\...
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Question: Differentiate $\tan ^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)$ with respect to $\sqrt{1-\mathrm{x}^{2}}$, if $-1x1$. Solution: Let $\mathrm{u}=\tan ^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)$ and $\mathrm{v}=\sqrt{1-\mathrm{x}^{2}}$ We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$. We have $u=\tan ^{-1}\left(\frac{1-x}{1+x}\right)$ By substituting $x=\tan \theta$, we have $\mathrm{u}=\tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\r...
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Question: Differentiate $\sin ^{-1}\left(2 \mathrm{ax} \sqrt{1-\mathrm{a}^{2} \mathrm{x}^{2}}\right)$ with respect to $\sqrt{1-\mathrm{a}^{2} \mathrm{x}^{2}}$, if $-\frac{1}{\sqrt{2}}\mathrm{ax}\frac{1}{\sqrt{2}}$ Solution: Let $u=\sin ^{-1}\left(2 a x \sqrt{1-a^{2} x^{2}}\right)$ and $v=\sqrt{1-a^{2} x^{2}}$. We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$. We have $u=\sin ^{-1}\left(2 a x \sqrt{1-a^{2} x^{2}}\right)$ $\Rightarrow \mathrm{u}=\sin ^{-1}\left(2 \ma...
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Question: Differentiate $\sin ^{-1} \sqrt{1-x^{2}}$ with respect to $\cot ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$, if $0x1$. Solution: Let $u=\sin ^{-1} \sqrt{1-x^{2}}$ and $v=\cot ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$ We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$. We have $u=\sin ^{-1} \sqrt{1-x^{2}}$ By substituting $x=\cos \theta$, we have $\mathrm{u}=\sin ^{-1} \sqrt{1-(\cos \theta)^{2}}$ $\Rightarrow u=\sin ^{-1} \sqrt{1-\cos ^{2} \theta}$ $\Right...
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Question: Differentiate $\tan ^{-1}\left(\frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}\right)$ with respect to $\sin ^{-1}\left(2 \mathrm{x} \sqrt{1-\mathrm{x}^{2}}\right)$, if $-\frac{1}{\sqrt{2}}\mathrm{x}\frac{1}{\sqrt{2}}$ Solution: Let $u=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$ and $v=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$. We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$. We have $u=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$ By substituting $x...
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Question: Differentiate $\cos ^{-1}\left(4 x^{3}-3 x\right)$ with respect to $\tan ^{-1}\left(\frac{1-x^{2}}{x}\right)$, if $\frac{1}{2}x1$. Solution: Let $u=\cos ^{-1}\left(4 x^{3}-3 x\right)$ and $v=\tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)$ We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$. We have $u=\cos ^{-1}\left(4 x^{3}-3 x\right)$ By substituting $x=\cos \theta$, we have $u=\cos ^{-1}\left(4 \cos ^{3} \theta-3 \cos \theta\right)$ But, $\cos 3 \theta=4...
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Question: Differentiate $\cos ^{-1}\left(4 x^{3}-3 x\right)$ with respect to $\tan ^{-1}\left(\frac{1-x^{2}}{x}\right)$, if $\frac{1}{2}x1$. Solution: Let $u=\cos ^{-1}\left(4 x^{3}-3 x\right)$ and $v=\tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)$ We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$. We have $u=\cos ^{-1}\left(4 x^{3}-3 x\right)$ By substituting $x=\cos \theta$, we have $u=\cos ^{-1}\left(4 \cos ^{3} \theta-3 \cos \theta\right)$ But, $\cos 3 \theta=4...
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Question: Differentiate $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ with respect to $\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$, if $-1x1$. Solution: Let $u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ and $v=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$. We need to differentiate $u$ with respect to $v$ that is find $\frac{\text { du }}{\text { dv }}$. We have $u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ By substituting $x=\tan \theta$, we have $\mathrm{u}=\sin ^{-1}\left(\frac{2 \tan \theta}{1...
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Question: Differentiate $\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$ with respect to $\sec ^{-1} x$. Solution: Let $u=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$ and $v=\sec ^{-1} x$ We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$. We have $u=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$ $\Rightarrow \mathrm{u}=\tan ^{-1}\left(\frac{\cos \left(2 \times \frac{\mathrm{x}}{2}\right)}{1+\sin \left(2 \times \frac{x}{2}\right)}\right)$ But, $\cos 2 \theta=\cos...
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Question: Differentiate $\tan ^{-1}\left(\frac{x-1}{x+1}\right)$ with respect to $\sin ^{-1}\left(3 x-4 x^{3}\right)$, if $-\frac{1}{2}x\frac{1}{2}$. Solution: Let $u=\tan ^{-1}\left(\frac{x-1}{x+1}\right)$ and $v=\sin ^{-1}\left(3 x-4 x^{3}\right)$ We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$. We have $\mathrm{u}=\tan ^{-1}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)$ By substituting $x=\tan \theta$, we have $\mathrm{u}=\tan ^{-1}\left(\frac{\tan \theta-1}{\...
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Question: Differentiate $\tan ^{-1}\left(\frac{2 \mathrm{x}}{1-\mathrm{x}^{2}}\right)$ with respect to $\cos ^{-1}\left(\frac{1-\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right)$, if $0\mathrm{x}1$. Solution: Let $u=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$ and $v=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$ We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$. We have $u=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$ By substituting $x=\tan \theta$, we have $\mathrm{u}=\tan ...
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Question: Differentiate $\sin ^{-1}\left(2 \mathrm{x} \sqrt{1-\mathrm{x}^{2}}\right)$ with respect to $\tan ^{-1}\left(\frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}\right)$, if $-\frac{1}{\sqrt{2}}\mathrm{x}\frac{1}{\sqrt{2}}$. Solution: Let $u=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ and $v=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$ We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$. We have $u=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ By substituting $x=\sin ...
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Question: Differentiate $\sin ^{-1}\left(2 \mathrm{x} \sqrt{1-\mathrm{x}^{2}}\right)$ with respect to $\tan ^{-1}\left(\frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}\right)$, if $-\frac{1}{\sqrt{2}}\mathrm{x}\frac{1}{\sqrt{2}}$. Solution: Let $u=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ and $v=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$ We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$. We have $u=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ By substituting $x=\sin ...
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Question: Differentiate $\tan ^{-1}\left(\frac{1+a x}{1-a x}\right)$ with respect to $\sqrt{1+a^{2} x^{2}}$. Solution: Let $u=\tan ^{-1}\left(\frac{1+\operatorname{ax}}{1-2 x}\right)$ and $v=\sqrt{1+a^{2} x^{2}}$ We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$. We have $u=\tan ^{-1}\left(\frac{1+a x}{1-a x}\right)$ By substituting $a x=\tan \theta$, we have $\mathrm{u}=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right)$ $\Rightarrow \mathrm{u}=\tan ^{-1}\l...
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