Derivative of $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ with respect to $\tan ^{-1} x$ for $-1 < x < 1$ is:

The derivative of $y = \sin^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)$ is

Differentiate the following with respect to $x$: $\sin ^{-1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$

If $y(x)=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$ and $\left(1+a^2 x^2\right) y^{\prime \prime}+g(x) y^{\prime}=0$, then the sum of the roots of the equation $1+a^2 x^2+g(x)=0$ is

If $y = \sin^{-1} \left( \frac{25 - x^2}{25 + x^2} \right)$, then find the value of $y'(1)$.

$\frac{d}{d x}\left(\cos ^{-1}\left(\frac{4 x^3}{27}-x\right)\right)=$