If $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}}$,then $f'(a) = $

Let $f$ and $g$ be differentiable functions on $R$ such that $f \circ g$ is the identity function. If for some $a, b \in R$, $g^{\prime}(a) = 5$ and $g(a) = b$, then $f^{\prime}(b)$ is equal to:

If the derivative of $\tan^{-1}(a + bx)$ with respect to $x$ at $x = 0$ is $1$, then $a^6 - b^3 = $

At $x = \sqrt{\frac{\pi}{2}}$,find the value of $\frac{d}{dx} \cos(\sin(x^2))$.

If $f(x) = \tan^{-1}\left( \frac{\sin x}{1 + \cos x} \right)$,then $f'\left( \frac{\pi}{3} \right) = $

$\frac{d}{dx} \left( \sqrt{3} \sin \left(2x + \frac{\pi}{3} \right) + \cos \left(2x + \frac{\pi}{3} \right) \right) = $ . . . . . .