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 $h(x) = \sqrt{4f(x) + 3g(x)}$,$f(1) = 4$,$g(1) = 3$,$f'(1) = 4$,and $g'(1) = 3$,then find $h'(1)$.

If $f(x) = x^2 - 3x$,then the points at which $f(x) = f'(x)$ are

The graph of function $f$ contains the points $P(1, 2)$ and $Q(s, r)$. The equation of the secant line through $P$ and $Q$ is $y = \left( \frac{s^2 + 2s - 3}{s - 1} \right) x - 1 - s$. The value of $f'(1)$ is:

The differential of ${e^{x^3}}$ with respect to $\log_e x$ is

If $\frac{d}{d x}\left[(x+1)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\right] = \left(15 x^p-16 x^q+1\right)(x-1)^{-2}$, then $(p, q)$ is equal to