Use the definition (1st principle) to compute f'(1).

Question

don't know how to deal with the absolute sign, please help $$f(x)=\sqrt{(|x^3 +1|)}$$ Use the definition ( first principle) to compute $f'(1)$

Show that $f(x)$ is not differentiable at $x=-1$

Answer 1

Derivative in $x=1$: you have to compute $ \lim_{x \to 1}\frac{f(x)-f(1)}{x-1}$. Since $x$ is near $1$ for $x \to 1$ , we can assume that $x>0$, hence $|x^3+1|=x^3+1$.

For the second part: show that $ \lim_{n \to \infty}\frac{f(-1+1/n)-f(-1)}{1/n}$ does not exist.