$$ \frac{d}{d x}\left[\operatorname{cosech}^{-1}(\tan 2 x)\right]= $$

Let $f: R \rightarrow R$ be defined by $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ for all $x$ and $y$. If $f^{\prime}(0)$ exists and equals -1 and $f(0)=1$, then $f(2)=$

If $\operatorname{Lt}_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=e^x(x+1)$ and $f(0)=0$, then $\frac{d}{d x}\left(f(x) e^{-x}\right)+\frac{d}{d x}\left(\frac{f(x)}{x}\right)=$

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