If $f(x)=\frac{2x}{4+3|x|}, x \in R$,then $f^{\prime}(0)=$

The function given by $y = ||x| - 1|$ is differentiable for all real numbers except the points

Consider the function $f:(0, \infty) \rightarrow \mathbb{R}$ defined by $f(x)=e^{-\left|\log _e x\right|}$. If $m$ and $n$ are respectively the number of points at which $f$ is not continuous and $f$ is not differentiable,then $m+n$ is

If $f(x) = \begin{cases} x^2 & \text{if } x \leqslant x_0 \\ ax + b & \text{if } x > x_0 \end{cases}$ is derivable for all $x \in \mathbb{R}$,then the values of $a$ and $b$ are respectively:

Let $f: R \rightarrow R$ be a function defined by $f(x)=\begin{cases} \frac{\sin(x^2)}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x=0 \end{cases}$. Then,at $x=0$,$f$ is

$f(x) = |\log_e |x||$ is differentiable at