Write the derivative of sin x with respect to cos x.
Question: Write the derivative of $\sin x$ with respect to $\cos x$. Solution: We have to find $\frac{d}{d(\cos x)}(\sin x)$ So, we use the Chain Rule of Differentiation to evaluate this. $\frac{d}{d(\cos x)}(\sin x)=\frac{d(\sin x)}{d x} \cdot \frac{d x}{d(\cos x)}$ $=\cos x \cdot \frac{1}{-\sin x}$ $=-\cot x($ Ans $)$...
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Question: If $f(x)$ is an even function, then write whether $f^{\prime}(x)$ is even or odd. Solution: $f(x)$ is an even function. This means that $\mathrm{f}(-\mathrm{x})=\mathrm{f}(\mathrm{x})$. If we differentiate this equation on both sides w.r.t. $x$, we get - $f^{\prime}(-x) \cdot(-1)=f^{\prime}(x)$ or, $-f^{\prime}(-x)=f^{\prime}(x)$ i.e., $f^{\prime}(x)$ is an odd function. (Ans)...
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Question: If $y=\log |3 x|, x \neq 0$, find $\frac{d y}{d x}$. Solution: $y=\log |3 x|$ So, $\frac{d y}{d x}=\frac{1}{3 x} \cdot 3$ $=\frac{1}{x}, x \neq 0$ i.e., $\frac{d y}{d x}=\frac{1}{x}, x \neq 0$ (Ans)...
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Question: If $f(x)=\log \left\{\frac{u(x)}{v(x)}\right\}, u(1)=v(1)$ and $u^{\prime}(1)=v^{\prime}(1)=2$, then find the value of $f^{\prime}(1)$. Solution: Using the Chain Rule of Differentiation, $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\frac{\mathrm{u}(\mathrm{x})}{\mathrm{v}(\mathrm{x})}} \cdot \frac{\mathrm{v}(\mathrm{x}) \cdot \mathrm{u}^{\prime}(\mathrm{x})-\mathrm{v}^{\prime}(\mathrm{x}) \cdot \mathrm{u}(\mathrm{x})}{(\mathrm{v}(\mathrm{x}))^{2}}$ $=\frac{v(x) \cdot u^{\prime}(x)-v^{\pri...
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Question: If $u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ and $v=\tan ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$, where $-1x1$, then write the value of $\frac{d u}{d v}$ Solution: $u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ and $v=\tan ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ We know, $\frac{\mathrm{du}}{\mathrm{dx}}=\frac{2}{1+\mathrm{x}^{2}}$ Using the chain rule of differentiation, $\frac{d v}{d x}=\frac{1}{1+\left(\frac{2 x}{1+x^{2}}\right)^{2}} \cdot \frac{\left(1+x^{2}\right) \cdot(2 x)^{\p...
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Question: If $|x|1$ and $y=1+x+x^{2}+\ldots$ to $\infty$, then find the value of $\frac{d y}{d x}$ Solution: Since $|x|1$, $y=1+x+x^{2}+\ldots$ to $\infty$ $=\frac{1}{1-x}$ $\therefore \frac{d y}{d x}=-\frac{1}{(1-x)^{2}} \cdot-1$ $=\frac{1}{(1-x)^{2}}($ Ans $)$...
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Question: If $y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)$, then write the value of $\frac{d y}{d x}$. Solution: $y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)$ $=\cos ^{-1}\left(\frac{x-1}{x+1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)$ Which exists for $-1 \leq \frac{x-1}{x+1} \leq 1$ and is equal to $\frac{\pi}{2}$ Now, $\frac{x-1}{x+1} \leq 1$ $\Rightarrow \frac{x-1}{x+1}-1 \leq 0$ $\Rightarrow \frac{x-1}{x+1}-\frac{x+...
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Question: If $y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$, find $\frac{d y}{d x}$ Solution: $-1\frac{1-x^{2}}{1+x^{2}} \leq 1$ holds for all $x \in \mathbb{R}$. So, $y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\frac{\pi}{2}$, for all $x \in \mathbb{R}$ $\left(\because \sin ^{-1} \mathrm{~m}+\cos ^{-1} \mathrm{~m}=\frac{\pi}{2}, \mathrm{~m} \in[-1,1]\right)$ Hence, $\frac{d y}{d x}=0$, for all...
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Question: If $\mathrm{y}=\log \sqrt{\tan \mathrm{x}}$, write $\frac{\mathrm{dy}}{\mathrm{dx}}$ Solution: This particular problem is a perfect way to demonstrate how simple but powerful the Chain Rule of Differentiation is. It is important to identify and break the problem into the individual functions with respect to which successive differentiation shall be done. In this case, this is the way to break down the problem - $\frac{d y}{d x}=\frac{d y}{d(\sqrt{\tan x})} \cdot \frac{d(\sqrt{\tan x})}...
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Question: if $y=\log _{a} x$, find $\frac{d y}{d x}$. Solution: $y=\log _{a} x=\frac{\log _{e} x}{\log _{e} a}$ $\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\log _{e} \mathrm{a}} \cdot \frac{1}{\mathrm{x}}$ $=\frac{1}{x \log _{e} a}$ (Ans)...
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Question: If $\mathrm{y}=\tan ^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)$, find $\frac{\mathrm{dy}}{\mathrm{dx}}$. Solution: $y=\tan ^{-1}\left(\frac{1-x}{1+x}\right)$ Using the Chain Rule of Differentiation, $\frac{d y}{d x}=\frac{1}{1+\left(\frac{1-x}{1+x}\right)^{2}} \cdot \frac{(1+x) \cdot(1-x)^{\prime}-(1+x)^{\prime} \cdot(1-x)}{(1+x)^{2}}$ $=\frac{(1+x)^{2}}{(1+x)^{2}+(1-x)^{2}} \cdot \frac{(1+x)(-1)-(1)(1-x)}{(1+x)^{2}}$ $=-\frac{2}{(1+x)^{2}+(1-x)^{2}}$ $=-\frac{1}{1+x^{2}}$ (An...
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Question: If $y=x^{x}$, find $\frac{d y}{d x}$ at $x=e$. Solution: $y=x^{x}$ Taking logarithm on both sides, $\log y=x \log x$ Differentiating w.r.t. $x$ on both sides, $\frac{1}{y} \cdot \frac{d y}{d x}=x \cdot \frac{1}{x}+1 \cdot \log x$ $=1+\log x$ $\Rightarrow \frac{d y}{d x}=y(1+\log x)$ $=x^{x}(1+\log x)$ So, at $x=e$, $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{e}}(1+\log \mathrm{e})$ $=e^{e}(1+1)$ $=2 e^{e}($ Ans $)$...
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Question: If $-\frac{\pi}{2}x0$ and $y=\tan ^{-1} \sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}$, find $\frac{d y}{d x} .$ Solution: $y=\tan ^{-1} \sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}$ $=\tan ^{-1} \sqrt{\frac{1-\left(1-2 \sin ^{2} x\right)}{1+\left(2 \cos ^{2} x-1\right)}}$ $=\tan ^{-1} \sqrt{\frac{2 \sin ^{2} x}{2 \cos ^{2} x}}$ $=\tan ^{-1} \sqrt{\tan ^{2} x}$ When $-\frac{\pi}{2}\mathrm{x}0, \tan \mathrm{x}$ is negative. So, square root of $\tan ^{2} \mathrm{x}$ in this condition is $-\tan \mathrm{x}...
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Question: If $x=a(\theta+\sin \theta), y=a(1+\cos \theta)$, find $\frac{d y}{d x}$ Solution: $\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(1+\cos \theta)$ and $\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(-\sin \theta)$ Using Chain Rule of Differentiation, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{d} \theta} \cdot \frac{\mathrm{d} \theta}{\mathrm{dx}}$ $=a(-\sin \theta) \cdot \frac{1}{a(1+\cos \theta)}$ $=-\frac{\sin \theta}{1+\cos \theta}$ $=-\frac{\sin \theta}{1+\cos ...
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Question: If $y=\sin ^{-1} x+\cos ^{-1} x$, find $\frac{d y}{d x}$. Solution: We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ So, here $y=\sin ^{-1} x+\cos ^{-1} x$ $=\frac{\pi}{2}$ which is a constant. Also, $\sin ^{-1} x$ and $\cos ^{-1} x$ exist only when $-1 \leq x \leq 1$ So, $\frac{d y}{d x}=0$ when $x \in[-1,1]$ and does not exist for all other values of $x$....
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Question: If $y=x|x|$, find $\frac{d y}{d x}$ for $x0$. Solution: $y=x|x|$ or, $y=\left\{\begin{array}{c}x^{2}, \text { when } x \geq 0 \\ -x^{2}, \text { when } x0\end{array}\right.$ So, for $x0$ $\frac{d y}{d x}=\frac{d}{d x}\left(-x^{2}\right)$ $=-2 \times($ Ans $)$...
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Question: If $f(0)=f(1)=0, f^{\prime}(1)=2$ and $y=f\left(e^{x}\right) e^{f(x)}$, write the value of $\frac{d y}{d x}$ at $x=0$. Solution: Using the Chain Rule of Differentiation, $\frac{d y}{d x}=u \cdot v^{\prime}+u^{\prime} \cdot v$ $=f\left(e^{x}\right) \cdot e^{f(x)} f^{\prime}(x)+f^{\prime}\left(e^{x}\right) e^{x} \cdot e^{f(x)}$ At $x=0$, $\frac{d y}{d x}=f\left(e^{0}\right) \cdot e^{f(0)} f^{\prime}(0)+f^{\prime}\left(e^{0}\right) e^{0} \cdot e^{f(0)}$ $=f(1) \cdot e^{f(0)} f^{\prime}(0)...
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Question: If $\mathrm{y}=\sin ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)$, write the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ for $\mathrm{x}1$ Solution: $y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=2 \tan ^{-1} x$ So, $\frac{\mathrm{dy}}{\mathrm{dx}}=2 \cdot \frac{1}{1+\mathrm{x}^{2}}$ $=\frac{2}{1+x^{2}}$ So, answer is $\frac{d y}{d x}=\frac{2}{1+x^{2}}$ (Ans)...
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Question: If $\pi \leq x \leq 2 \pi$ and $y=\cos ^{-1}(\cos x)$, find $\frac{d y}{d x}$ Solution: $y=\cos ^{-1}(\cos x)$ for $x \in(\pi, 2 \pi)$ $y=\cos ^{-1}(\cos x)$ $=\cos ^{-1}(\cos (\pi+(x-\pi)))$ $=\cos ^{-1}(-\cos (x-\pi))$ $=\pi-(x-\pi)$ $=2 \pi-x$ [Since, $\cos (\pi+x)=-\cos x$ and $\left.\cos ^{-1}(-x)=\pi-x\right]$ So, $\frac{\mathrm{dy}}{\mathrm{dx}}=-1$ For $\cos ^{-1}(\cos x), x=n_{\pi}$ are the 'sharp corners' where slope changes from 1 to $-1$ or vice versa, i.e., the points wher...
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Question: If $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}$ and $y=\sin ^{-1}(\sin x)$, find $\frac{d y}{d x}$ Solution: For $\mathrm{x} \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ $y=\sin ^{-1}(\sin x)$ $=\sin ^{-1}(\sin (\pi-(\pi-x))$ (to get $y$ in principal range of $\sin ^{-1} x$ ) i.e., $y=\pi-x$ $\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=-1$ From the last problem we see that $\frac{d y}{d x}_{x \rightarrow \frac{\pi^{-}}{2}}=1$ and $\frac{d y}{d x}_{x \rightarrow \frac{\pi^{+}}{2}}=-1$ ...
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Question: If $y=\sin ^{-1}(\sin x),-\frac{\pi}{2} \leq x \leq \frac{\pi}{2} .$ Then write the value of $\frac{d y}{d x}$ for $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ Solution: For $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ $y=\sin ^{-1}(\sin x)$ $=x$ So, $\frac{d y}{d x}=1$ (Ans)...
Read More →Let g(x) be the inverse of an invertible function
Question: Let $g(x)$ be the inverse of an invertible function $f(x)$ which is derivable at $x=3 .$ If $f(3)=9$ and $f^{\prime}(3)=9$, write the value of $g^{\prime}(9)$. Solution: From the definition of invertible function, $g(f(x))=x \ldots$ (i) So, $g(f(3))=3$, i.e., $g(9)=3$ Now, differentiating both sides of equation (i) w.r.t. $x$ using the Chain Rule of Differentiation, we get - $g^{\prime}(f(x)), f^{\prime}(x)=1 \ldots($ ii $)$ Plugging in $x=3$ in equation (ii) gives us - $g^{\prime}(f(3...
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Question: If $f^{\prime}(x)=\sqrt{2 x^{2}-1}$ and $y=f\left(x^{2}\right)$, then find at $x=1$. Solution: $y=f\left(x^{2}\right)$ $\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}\left(\mathrm{x}^{2}\right) \cdot 2 \mathrm{x}$ $=2 x \sqrt{2\left(x^{2}\right)^{2}-1}$ $=2 x \sqrt{2 x^{4}-1}$ Putting $x=1$, $\frac{d y}{d x}=2 \cdot 1 \cdot \sqrt{2 \cdot 1^{4}-1}$ $=2 \sqrt{2-1}$ $=2 \sqrt{1}$ $=2$ i.e., $\frac{\mathrm{dy}}{\mathrm{dx}}=2$ at $\mathrm{x}=1$. (Ans)...
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Question: If $f(1)=4, f^{\prime}(1)=2$, find the value of the derivative of $\log \left(f\left(e^{x}\right)\right)$ with respect to $x$ at the point $x=0$. Solution: Using the Chain Rule of Differentiation, derivative of $\log \left(f\left(e^{x}\right)\right)$ w.r.t. $x$ is $\frac{1}{f\left(e^{x}\right)} \cdot f^{\prime}\left(e^{x}\right)$ So, the value of the derivative at $x=0$ is $\frac{1}{f\left(e^{0}\right)} \cdot f^{\prime}\left(e^{0}\right)=\frac{1}{f(1)} \cdot f^{\prime}(1)$ $=\frac{1}{4...
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Question: If $f^{\prime}(1)=2$ and $y=f\left(\log _{e} x\right)$, find $\cdot \frac{d y}{d x} \cdot$ at $x=e$ Solution: $y=f\left(\log _{e} x\right)$ Using the Chain Rule of Differentiation, $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}\left(\log _{\mathrm{e}} \mathrm{x}\right) \cdot \frac{1}{\mathrm{x}}$ So, at $\mathrm{x}=\mathrm{e}$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}\left(\log _{\mathrm{e}} \mathrm{e}\right) \cdot \frac{1}{\mathrm{e}}$ $=f^{\prime}(1) \cdot \frac{1}{e...
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