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

If the function $f(x) = \begin{cases} \frac{k\cos x}{\pi - 2x} & : x \neq \frac{\pi}{2} \\ 3 & : x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$, then $k$ is equal to

If $y = t - \frac{1}{t}$ and $x = t + \frac{1}{t}$, then $\frac{dy}{dx}$ is equal to

Let $y = \log_e \left(\frac{a + b \sin x}{a - b \sin x}\right)$, then value of $\frac{dy}{dx}$ is :

Let $e^y(x+1) = 1$. Then which of the following are TRUE?

(A) $\frac{d^2y}{dx^2} = -\frac{1}{(x+1)^2}$

(B) $\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$

(C) $\left.\frac{d^2y}{dx^2}\right|_{x=0} = -1$

(D) $\left.\frac{d^2y}{dx^2}\right|_{x=0} = 1$

(E) $\left.\frac{d^2y}{dx^2}\right|_{x=1} = \frac{1}{4}$

Choose the correct answer from the options given below:

If $y = \log_e \sin(e^x + 5x + 500)$ then $\dfrac{dy}{dx}$ is equal to

If $y = \frac{1}{1+x^{b-a}+x^{c-a}} + \frac{1}{1+x^{c-b}+x^{a-b}} + \frac{1}{1+x^{a-c}+x^{b-c}}$ then $\frac{d^2y}{dx^2}$ is