Let $f(x) = x - \frac{1}{x}$,then $f^{\prime}(-1)$ is

If $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}}$, then $f^{\prime}(a)$ is equal to

If $f(x) = \frac{x}{1+|x|}$,for $x \in \mathbb{R}$,then $f^{\prime}(0)$ is equal to :

If $y(\alpha)=\sqrt{2\left(\frac{\tan \alpha+\cot \alpha}{1+\tan ^{2} \alpha}\right)+\frac{1}{\sin ^{2} \alpha}}$ for $\alpha \in\left(\frac{3 \pi}{4}, \pi\right)$,then find $\frac{d y}{d \alpha}$ at $\alpha=\frac{5 \pi}{6}$.

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

Differentiate the function with respect to $x$: $\sin^{-1}(x\sqrt{x})$,where $0 \le x \le 1$.