If $y = \frac{1}{x+1}$, then $\frac{d^2y}{dx^2}$ at $x = 2$ is:

The value of k for which the function, defined by, $f(x) = \begin{cases} \frac{3x + 4 \tan x}{x} & : x \neq 0 \\ k & : x = 0 \end{cases}$ is continuous at $x = 0$, is

For the function $f(x) = \sqrt{x}$, which of the following statements are true?

(A) $f(x)$ is continuous for $x \geq 0$.

(B) $f(x)$ is continuous for $x \in R$ (Set of real numbers).

(C) $f(x)$ is differentiable for $x > 0$.

(D) $f(x)$ is differentiable for $x \in R$ (Set of real numbers).

Choose the correct answer from the options given below:

If $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ then $\frac{dy}{dx} =$

For the function $f(x) = -2|\log_e|x||$, which of the following is true?

The derivative of $(\log x)^x$ with respect to $\log x$ is

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: