Question

Differentiate ${\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ w.r.t ${\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$

Answer 1

We have

${\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\text{w}\text{.r}\text{.t}{\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$

Let put

$x=\cos \theta \text{,Then}\theta ={\cos }^{-1}x$

$\text{So,}$

$\text{ta}{\text{n}}^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right){\sec }^{-1}\left(\frac{1}{\sqrt{2x^2-x1}}\right)$

${\tan }^{-1}\left(\frac{\cos \theta }{\sqrt{1-{\cos }^2\theta }}\right){\sec }^{-1}\left(\frac{1}{\sqrt{2{\cos }^2\theta -1}}\right)$

$\Rightarrow {\tan }^{-1}\cot \theta {\sec }^{-1}\sec 2\theta$

$\Rightarrow \theta ,2\theta$

${\cos }^{-1}x,2{\cos }^{1}x$

$\text{let}$

$\text{u}\text{=}\text{co}{\text{s}}^{-1}x,v=2{\cos }^{-1}x$

On. Diff. with respect to x,

$\frac{du}{dx}=\frac{-1}{\sqrt{1-x^2}},\frac{dv}{d}=\frac{-2}{\sqrt{1-x^2}}$

According to given question.

$\frac{du}{dv}=\frac{\frac{1}{\sqrt{1-x^2}}}{\frac{-2}{\sqrt{1-x^2}}}$

$\frac{du}{dv}=\frac{1}{2}$

Hence, this is the answer.