Differentiating trigonometric function with respect to another trigonometric function

Question

I am unable to get even a Lil bit of clue of How to differentiate

with respect to

Can I get some support here... I am tryin' to figure this math out... But I lack a few techniques.

Answer 1

Let the $2$ functions be $u(x)$ and $v(x)$.
Then $$\frac{du(x)}{dv(x)} = \frac{du(x)/dx}{dv(x)/dx}$$

In this case $u(x) =\cos^{-1}\frac{1-x^2}{1+x^2} $ and $v(x) = \tan^{-1}\frac{2x}{1-x^2}$

Answer 2

Hint:

Let $z=\arctan x,-\dfrac\pi2<z<\dfrac\pi2$ and $x=\tan z$

$$\cos^{-1}\dfrac{1-x^2}{1+x^2}=\cos^{-1}(\cos2z)=\begin{cases}2z &\mbox{if }0\le2z\le\pi \\-2z & \mbox{if }0\le-2z\le\pi\iff-\pi\le2z\le0 \end{cases}$$

So, $$\dfrac{d\cos^{-1}(\cos2z))}{dz}=?$$

Set $y=x$ in Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$