If $\sqrt{1-x^2} + \sqrt{1-y^2} = a(x-y)$, then $\frac{dy}{dx} =$

If $t=e^{2 x}$ and $y=\log _{e} t^{2}$, then $\frac{d^{2} y}{d x^{2}}$ is :

If the function $f(x) = \dfrac{(4^x-1)^3}{\sin\dfrac{x}{4}\cdot\log\left(1+\dfrac{x^2}{3}\right)}$ is continuous at $x = 0$, then the value of $f(0)$ is: (Take $\log_e x = \log x$)

If $y = \left(x + \sqrt{x^2+1}\right)^m$, then $\frac{dy}{dx}$ 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 $f(x) = \begin{cases} ax^2 + b, & x < -1 \\ bx^2 + ax + 4, & x \geq -1 \end{cases}$ is everywhere differentiable, then :

If $y = (1+x)(1+x^2)(1+x^4)$, then $\frac{dy}{dx}$ at $x = -1$ is