Find the derivative of $y=\tan^{-1}\left(\frac {1}{1-x^2}\right)$

Question

Find the derivative of $y=\tan^{-1}\left(\dfrac {1}{1-x^2}\right)$

My Approach $$y=\tan^{-1} \left(\dfrac {1}{1-x^2}\right)$$ Let $x=\sin (\theta)$ Then, $$y=\tan^{-1} \left(\dfrac {1}{1-\sin^2 (\theta)}\right)$$ $$y=\tan^{-1} \left(\dfrac {1}{\cos^2 (\theta)}\right)$$ $$y=\tan^{-1} (\sec^2 (\theta))$$

How do I proceed further?

Answer 1

Since $\cot y=1-x^2$,

\begin{align*} -\csc^2y\frac{dy}{dx}&=-2x\\ -(1+\cot^2y)\frac{dy}{dx}&=-2x\\ \frac{dy}{dx}&=\frac{2x}{1+(1-x^2)^2}\\ &=\frac{2x}{x^4-2x^2+2} \end{align*}

Answer 2

Why not using the chain rule instead:

$$ y' = \frac{1}{1 + (1/(1-x^2))^2} \times \frac{ d }{d x} \left( \frac{1}{1-x^2} \right) = \frac{(1-x^2)^2}{1+(1-x^2)^2} \times \frac{2x}{(1-x^2)^2}$$

Answer 3

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

Derivating both sides we get:

$y'.\frac{1}{1+\frac{1}{(1-x^2)^2}}=(\frac{1}{1-x^2})' ⇒ y'=\frac{(1-x^2)^2}{1+(1-x^2)^2}\times \frac{2x}{(1-x^2)^2}$