A function $f(x) = \begin{cases} \frac{kx^3}{4} & \text{if } x \leq 1 \\ 8 & \text{if } x > 1 \end{cases}$ for some constant $k$, then $\frac{d^2f(x)}{dx^2}$ at $x = -1$ is

If $y = \sqrt{x + \sqrt{x + \sqrt{x + ...\ ...\ ...}}}$, then

If $y=e^{2\log_e t}$ and $x=\log_e e^{t^2}$, then the value of $\frac{d^2y}{dx^2}$ is

If $e^y(1+x)=1$, then $\dfrac{d^2y}{dx^2}$ is:

If $f(x) = \begin{cases} \frac{x^2+3x-10}{x-2}, & x \neq 2 \\ 2k+1, & x = 2 \end{cases}$ is continuous at $x = 2$, then the value of $k$ is

If $y=\sqrt[5]{\log_e(5x^2)}$, then $\dfrac{dy}{dx}=$

(Take $\log_e x=\log x$)

If $x = a\sin 2t(1 + \cos 2t)$ and $y = b\cos 2t(1 - \cos 2t)$, then $(\frac{dy}{dx})_{\text{at } x=\frac{\pi}{4}}$ is equal to