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

$\cos ^{-1}\dfrac {1-x^{2}}{1+x^{2}}$

with respect to

$\tan ^{-1}\dfrac {2x}{1-x^{2}}$

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

What does it mean to differentiate one function with respect to another? I am not entirely sure what you mean. Rewording the problem might help everybody. — Matthew Leingang, Feb 04 '20 at 17:50
Since you repeated the phrase, I'm going to repeat the question: what does that mean? — Matthew Leingang, Feb 04 '20 at 17:54

score 1 · Accepted Answer · answered Feb 04 '20 at 17:53

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 $v(x) = \tan^{-1}\frac{2x}{1-x^2}$

answered Feb 04 '20 at 17:53

19aksh

12,768

Yes... this is the thing... thank God you got what I meant to say!!! – Jency Feb 04 '20 at 17:54
1

You're welcome! – 19aksh Feb 04 '20 at 17:55
1

You can further deduce that $\cos^{-1}\frac{1-x^2}{1+x^2} = 2\tan^{-1}x$ and $\tan^{-1}\frac{2x}{1-x^2}=2\tan^{-1}x$ under certain conditions. – 19aksh Feb 04 '20 at 17:59
1

YES..... I remember that formula...Appreciate your co-operation, Boss @Ak19 – Jency Feb 04 '20 at 18:01
1

The result is 1... I solved it... – Jency Feb 04 '20 at 18:09

score 1 · Answer 2 · answered Feb 05 '20 at 09:46

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 $xy>1$

Differentiating trigonometric function with respect to another trigonometric function

2 Answers2