If $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), 0 < x < 1$, then $\frac{dy}{dx}$ is equal to

Whenever you see $\frac{1-x^2}{1+x^2}$, it should remind you of the identity:

$\cos 2\theta = \dfrac{1 - \tan^2\theta}{1 + \tan^2\theta}$

So we substitute $x = \tan\theta$, which means $\theta = \tan^{-1}x$.

$y = \cos^{-1}\!\left(\dfrac{1 - x^2}{1 + x^2}\right)$

Putting $x = \tan\theta$:

$y = \cos^{-1}\!\left(\dfrac{1 - \tan^2\theta}{1 + \tan^2\theta}\right)$

$= \cos^{-1}(\cos 2\theta)$

Since $0 < x < 1$ and $x = \tan\theta$, we get $0 < \theta < \dfrac{\pi}{4}$, so $0 < 2\theta < \dfrac{\pi}{2}$.

This lies within $[0, \pi]$, which is the range of $\cos^{-1}$, so:

$\cos^{-1}(\cos 2\theta) = 2\theta$

Therefore:

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

Differentiating with respect to $x$:

$\dfrac{dy}{dx} = 2 \cdot \dfrac{1}{1 + x^2}$

$= \dfrac{2}{1 + x^2}$

Useful substitution patterns to remember:

$\dfrac{1-x^2}{1+x^2}$ with $x = \tan\theta$ simplifies to $\cos 2\theta$

$\dfrac{2x}{1+x^2}$ with $x = \tan\theta$ simplifies to $\sin 2\theta$

$\dfrac{2x}{1-x^2}$ with $x = \tan\theta$ simplifies to $\tan 2\theta$