The derivative of $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$ with respect to $\sin ^{-1}\left(3 x-4 x^3\right)$ is

If $y=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$,where $0 \leq x < \frac{\pi}{2}$,then find the value of $\frac{d y}{d x}$ at $x=\frac{\pi}{6}$.

$\frac{d}{d x} \tan ^{-1}\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right]$ is equal to

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 $x=0$,$\frac{d}{d x}\left(f\left(\alpha e^{2 x}\right)\right)=$

If $f'(x) = \sin^2 x$ and $y = f \left( \frac{2x - 1}{x^2 + 1} \right)$, then $\frac{dy}{dx}$ at $x = 1$ is

If $y=\tan ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right\}$,then find $\frac{d y}{d x}$.