$\frac{d}{dx} \left( \frac{2 \tan x}{1 + \tan^2 x} \right) = $ . . . . . .

People living on Mars,instead of the usual definition of derivative $D f(x)$,define a new kind of derivative,$D^*f(x)$ by the formula $D^*f(x) = \lim_{h \to 0} \frac{f^2(x + h) - f^2(x)}{h}$ where $f^2(x)$ means $[f(x)]^2$. If $f(x) = x \ln x$,then the value of $\left. D^*f(x) \right|_{x = e}$ is:

If $f$ is differentiable, $f(x+y)=f(x) f(y)$ for all $x, y \in R$, $f(3)=3$, and $f^{\prime}(0)=11$, then $f^{\prime}(3)$ is equal to:

If $y = \sinh^{-1}\left(\frac{1-x}{1+x}\right)$,then find $\frac{dy}{dx}$.

If $f(x)=\operatorname{cosec}^{-1}\left[\frac{10}{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}\right]$,then $f^{\prime}(x)$ is equal to:

$\frac{d}{d x}\left[\cos ^2\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right]$ is