If the function $f(x)=\begin{cases} x^2, & x\le c \\ ax+b, & x>c \end{cases}$ is differentiable at $x=c$, then the relation between $b$ and $c$ is:

If $y = e^{acos^{-1}x}, -1 < x < 1$, then $(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx}$ is equal to

Derivative of $x^3 + 1$ with respect to $x^2 + 1$ is

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 $y = 5e^{2x} + 4e^{3x}$, then $\frac{d^2y}{dx^2}$ equals:

Match List-I with List-II

$\begin{array}{|l|l|} \hline \rule{0pt}{2.8ex}\text{List-I} & \text{List-II} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(A) } f(x) = |x| & \text{(I) Not differentiable at } x=-2 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(B) } f(x) = |x+2| & \text{(II) Not differentiable at } x=0 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(C) } f(x) = |x^2-4| & \text{(III) Not differentiable at } x=2 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(D) } f(x) = |x-2| & \text{(IV) Not differentiable at } x=2,-2 \text{ only} \\[1.2ex] \hline \end{array}$

Choose the correct answer from the options given below:

If $x = t^3$, $y = t^2$ then $\frac{d^2y}{dx^2}$ is equal to: