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

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

If $y=\frac{\log x}{x}$, then $\frac{d^2y}{dx^2}$ is equal to(Consider $\log_e x = \log x$):

If $y = \left[\dfrac{2}{1+x^{b-a}+x^{c-a}} + \dfrac{2}{1+x^{c-b}+x^{a-b}} + \dfrac{2}{1+x^{a-c}+x^{b-c}}\right]^x$, then $\dfrac{dy}{dx}$ is equal to :

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 $y = \sqrt{ax + b}$ then $y\frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 =$

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