$\frac{d}{dx}(\sin 2x^2)$ equals

$\frac{d}{dx}[\sin^n x \cos nx] = $

If $f(x) = \begin{cases} x-5, & \text{for } x \leq 1 \\ 4x^2-9, & \text{for } 1 < x < 2 \\ 3x+4, & \text{for } x \geq 2 \end{cases}$,then $f^{\prime}(2^{+})$ is equal to:

If $y = (x^x)^x$,then $\frac{dy}{dx} = $

For the differentiable function $f: R - \{0\} \rightarrow R$,let $3 f(x) + 2 f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$. Then $\left|f(3) + f^{\prime}\left(\frac{1}{4}\right)\right|$ is equal to

If $y = \sin^{-1}\sqrt{1 - x} + \cos^{-1}\sqrt{x}$,then $\frac{dy}{dx} = $