Suppose $f(x)=x(x+3)(x-2)$,where $x \in [-1,4]$. Then,a value of $c$ in $(-1,4)$ satisfying $f^{\prime}(c)=10$ is

If $f(x) = \frac{1}{1 + \frac{1}{x}}$ and $g(x) = \frac{1}{1 + \frac{1}{f(x)}}$, then $g^{\prime}(2)$ is equal to

Differentiate $a^{x}$ with respect to $x$,where $a$ is a positive constant.

If $G(x) = -\sqrt{25 - x^2}$,then $\mathop{\lim}\limits_{x \to 1} \frac{G(x) - G(1)}{x - 1} = $

Let $f: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$ be a differentiable function such that $f(0)=\frac{1}{2}$. If $\lim _{x \rightarrow 0} \frac{x \int_0^x f(t) dt}{e^{x^2}-1}=\alpha$,then $8 \alpha^2$ is equal to:

$\frac{d}{dx}\{ \log (\sec x + \tan x)\} = $