If $x = a(\theta+\sin\theta)$, $y = a(1-\cos\theta)$ then $\dfrac{d^2y}{dx^2}$ is equal to:

The condition on a and b, such that for $y = \frac{a}{x} - \frac{b}{x^2}$, $\frac{dy}{dx} = 0$ at $x=1$ is :

If $x\sqrt{1+y}+y\sqrt{1+x}=0, x\neq y$, then $\frac{dy}{dx}$ is equal to

Match List-I with List-II. Here [x] denotes the greatest integer function

$\begin{array}{|l|l|} \hline \rule{0pt}{2.8ex}\text{List-I} & \text{List-II} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(A) } f(x) = [x] & \text{(I) is continuous everywhere but not differentiable at } x=-1 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(B) } f(x) = |x-1| & \text{(II) is continuous everywhere except at all integral values} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(C) } f(x) = e^{|x|} & \text{(III) is continuous everywhere but not differentiable at } x=1 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(D) } f(x) = |x+1| & \text{(IV) is continuous everywhere but not differentiable at } x=0 \\[1.2ex] \hline \end{array}$

Choose the correct answer from the options given below:

The derivative of $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ w.r.t. $\tan^{-1}\left(\frac{2x}{1-x^2}\right)$ is

If $x=\log_e t$ and $y=\dfrac{1}{t}$, then $\dfrac{d^2y}{dx^2}$ at $t=0.1$ is

If $y = \sin^{-1} x + \sin^{-1} \sqrt{1-x^2}, x \in (-1, 0)$, then $\frac{dy}{dx}$ is equal to