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
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
- $\frac{1}{2} \sqrt{1-x^2} x^{\tan ^{-1} x}\left[\frac{\log x}{1+x^2}+\frac{\tan ^{-1} x}{x}\right]$
- $-\frac{1}{2} \sqrt{1-x^2} x^{x^{-a^{-1}}}\left[\log \left(\tan ^{-1} x\right)+x\left(1+x^2\right) \tan ^{-1} x\right]$
- $\frac{\left.-2 \tan ^{-1} \frac{\log x}{1+x^2}+\frac{\tan ^{-1} x}{x}\right]}{\sqrt{1-x^2}}$
- $-\frac{1}{2} \sqrt{1-x^2} x^{\tan x^{-1}}\left[\frac{\log x}{1+x^2}+\frac{\tan ^{-1} x}{x}\right]$
Solution
We have,
$
\begin{aligned}
f(x) & =x^{\tan ^{-1} x} \\
\Rightarrow \quad \log f(x) & =\tan ^{-1} x \log x
\end{aligned}
$
$\begin{aligned} & \therefore \quad \frac{1}{f(x)} \cdot \frac{d}{d x} f(x)=\frac{1}{1+x^2} \log x+\frac{\tan ^{-1} x}{x} \\ & \Rightarrow \quad \frac{d f(x)}{d x}=x^{\tan ^{-1} x}\left[\frac{\log x}{1+x^2}+\frac{\tan ^{-1} x}{x}\right] \\ & \text { Also, } \quad g(x)=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right) \\ & =\cos ^{-1}\left(2 x^2-1\right)=2 \cos ^{-1} x \\ & \therefore \quad \frac{d}{d x} g(x)=\frac{-2}{\sqrt{1-x^2}} \\ & \therefore \quad \frac{d f(x)}{d \cdot g(x)}=\frac{\frac{d f(x)}{d x}}{\frac{d g(x)}{d x}} \\ & =\frac{x^{\tan ^{-1} x}\left[\frac{\log x}{1+x^2}+\frac{\tan ^{-1} x}{x}\right]}{\frac{-2}{\sqrt{1-x^2}}} \\ & =\frac{-1}{2} \sqrt{1-x^2} \cdot x^{\tan ^{-1} x}\left[\frac{\log x}{1+x^2}+\frac{\tan ^{-1} x}{x}\right] \\ & \end{aligned}$
Asked in: AP EAMCET 2018 (22 Apr Shift 1)
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