For $f:[-1,1]\longrightarrow \mathbb{R}, f(x) =\arcsin(1-2x^2)$. Check if $f$ is differentiable at $\boldsymbol{x = 0}$

Question

For $f:[-1,1]\longrightarrow \mathbb{R}, f(x) = \arcsin(1-2x^2)$. Check if $f$ is differentiable at $x = 0$.

I've tried with l'Hospital but it didn't help me so far.

Answer 1

No, it isn't. Note that$$f'(x)=-\frac{2x}{\sqrt{x^2-x^4}}=\begin{cases}-\frac2{\sqrt{1-x^2}}&\text{ if }x>0\\\frac2{\sqrt{1-x^2}}&\text{ if }x<0.\end{cases}$$Therefore, the right derivative of $f$ at $0$ is $-2$, whereas the left derivative at $0$ is $2$.

Answer 2

Try $$\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$$ for $$x_0=0$$