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

If $e^y(x + 1) = 1$, then

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

If $f(x)$, defined by $f(x)=\left\{\begin{array}{lll}k x+1 & \text { if } & x \leq \pi \\ \cos x & \text { if } & x>\pi\end{array}\right.$ is continuous at $x=\pi$, then the value of $k$ is :

The point('s) at which the function $f$ given by $f(x) = \begin{cases} \dfrac{|x|}{2x}, & x<0 \\ -\dfrac{1}{2}, & x\geq0 \end{cases}$ is continuous is (are)

If $y = (log x)^{(log x)}$, $x > 1$ then $\frac{dy}{dx}$ is equal to

Match List-I with List-II

Choose the correct answer from the options given below: