The function $y = \sin^{-1}(\cos x)$ is not differentiable at . . . . . .

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 $f(x) = \begin{cases} x \left(1 + \frac{1}{2} \sin (\log x^2) \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$,then find the value of $\lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x}$.

Suppose $f(x) = \begin{cases} [\cos \pi x], & x \leq 1 \\ 2\{x\} - 1, & x > 1 \end{cases}$, where $[\cdot]$ and $\{\cdot\}$ denote the greatest integer function and the fractional part of $x$ respectively, then at $x = 1$:

If $f(x + y) = f(x) + f(y) + |x|y + xy^2$,$\forall x, y \in R$ and $f'(0) = 0$,then

If $f(x) = \begin{cases} \tan^{-1} x, & \text{when } |x| \leq 1 \\ \frac{1}{2}(|x|-1), & \text{when } |x| > 1 \end{cases}$,then the domain of $\frac{d}{dx} f(x)$ is