If $y = \sin^{-1}\left( \frac{1 - x^2}{1 + x^2} \right)$,then $\frac{dy}{dx}$ equals

Let $y=f(x)=\sin ^3\left(\frac{\pi}{3}\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^3+5 x^2+1\right)^{\frac{3}{2}}\right)\right)$. Then,at $x =1$,

$\frac{d}{dx} \left( \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right) \right)$ is equal to

If $y = \tan^{-1} \left( \frac{3\cos x - 4\sin x}{4\cos x + 3\sin x} \right) + 2\tan^{-1} \left( \frac{x}{1+\sqrt{1-x^2}} \right)$,then $\frac{dy}{dx}$ at $x = \frac{\sqrt{3}}{2}$ is equal to:

Differentiate $\tan ^{-1} x$ with respect to $\cot ^{-1} x$ for $x \in R$.

Let $f: R \rightarrow R$ be a continuous function. If $px+my+n=0$ is a tangent drawn to the curve $y=f(x)$ at $x=\alpha$,then at $x=0$,$\frac{d}{d x}\left(f\left(\alpha e^{2 x}\right)\right)=$