Question

Differentiate the function ${\tan }^{-1}\frac{2x}{1-x^2}$ w.r.to ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

Answer 1

Let $\theta ={\tan }^{-1}\frac{2x}{1-x^2}$

$\therefore \tan \theta =\frac{2x}{1-x^2}$

Substitute $x=\tan \alpha$

$\therefore \tan \theta =\frac{2\tan \alpha }{1-{\tan }^2\alpha }$

$\therefore \tan \theta =\tan 2\alpha$

$\therefore \theta =2\alpha$ (considering only primary values)

$\therefore \theta =2{\tan }^{-1}x$

Differentiating wrt $x$, we get

$\therefore \frac{d\theta }{dx}=\frac{2}{1+x^2}$

Now, consider $\gamma ={\cos }^{-1}\frac{1-x^2}{1+x^2}$

$\therefore \cos \gamma =\frac{1-x^2}{1+x^2}$

Substitute $x=\tan \beta$

$\therefore \cos \gamma =\frac{1-{\tan }^2\beta }{1+{\tan }^2\beta }$

$\therefore \cos \gamma =\frac{1-\frac{{\sin }^2\beta }{{\cos }^2\beta }}{1+\frac{{\sin }^2\beta }{{\cos }^2\beta }}$

$\therefore \cos \gamma =\frac{\frac{{\cos }^2\beta -{\sin }^2\beta }{{\cos }^2\beta }}{\frac{{\cos }^2\beta +{\sin }^2\beta }{{\cos }^2\beta }}$

$\therefore \cos \gamma =\frac{\cos 2\beta }{1}$

(as ${\cos }^2\beta +{\sin }^2\beta =1$ and ${\cos }^2\beta -{\sin }^2\beta =\cos 2\beta$)

$\therefore \cos \gamma =\cos 2\beta$

$\therefore \gamma =2\beta$ (considering only primary values)

$\therefore \gamma =2{\tan }^{-1}x$

Differentiating wrt $x$, we get

$\therefore \frac{d\gamma }{dx}=\frac{2}{1+x^2}$

Now,

$\frac{d\left\{{\tan }^{-1}\frac{2x}{1-x^2}\right\}}{d\left\{{\cos }^{-1}\frac{1-x^2}{1+x^2}\right\}}=\frac{d\theta }{d\gamma }$

$=\frac{d\theta }{dx}\times \frac{dx}{d\gamma }$

$=\frac{2}{1+x^2}\times \frac{1+x^2}{2}$

$=1$

This is the required solution.