The derivative of f(x)=x^{tan ^{-1} x} with r | vedclass

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Question

(A) Given $f(x) = x^{	an ^{-1} x}$. Taking logarithm on both sides,$\log f(x) = 	an ^{-1} x \cdot \log x$. Differentiating with respect to $x$,$\fra

Vedclass · Accepted Answer

$-\frac{1}{2} \sqrt{1-x^2} x^{	an ^{-1} x}\left[\frac{\log x}{1+x^2}+\frac{	an ^{-1} x}{x}ight]$

Vedclass · Answer

(A) Given $f(x) = x^{	an ^{-1} x}$. Taking logarithm on both sides,$\log f(x) = 	an ^{-1} x \cdot \log x$. Differentiating with respect to $x$,$\frac{1}{f(x)} \frac{df}{dx} = \frac{1}{1+x^2} \log x + \frac{	an ^{-1} x}{x}$. Thus,$\frac{df}{dx} = x^{	an ^{-1} x} \left[ \frac{\log x}{1+x^2} + \frac{	an ^{-1} x}{x} ight]$. Now,$g(x) = \sec ^{-1} \left( \frac{1}{2x^2-1} ight) = \cos ^{-1} (2x^2-1)$. Using the substitution $x = \cos 	heta$,$g(x) = \cos ^{-1} (2 \cos^2 	heta - 1) = \cos ^{-1} (\cos 2	heta) = 2	heta = 2 \cos ^{-1} x$. Differentiating $g(x)$ with respect to $x$,$\frac{dg}{dx} = 2 \cdot \left( -\frac{1}{\sqrt{1-x^2}} ight) = -\frac{2}{\sqrt{1-x^2}}$. Finally,the derivative of $f(x)$ with respect to $g(x)$ is $\frac{df}{dg} = \frac{df/dx}{dg/dx} = \frac{x^{	an ^{-1} x} \left[ \frac{\log x}{1+x^2} + \frac{	an ^{-1} x}{x} ight]}{-2/\sqrt{1-x^2}} = -\frac{1}{2} \sqrt{1-x^2} x^{	an ^{-1} x} \left[ \frac{\log x}{1+x^2} + \frac{	an ^{-1} x}{x} ight]$.

The derivative of $f(x)=x^{\tan ^{-1} x}$ with respect to $g(x)=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ is

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