How to find $f'(0)$

Question

Let $f(x)$ be a function satisfying $|f(x)| \le x^2$ for $-1 \le x \le 1$. Show that $f$ is differentiable at $x=0$ and find $f'(0)$.

Show that $$f(x)=\begin{cases}x^2 \sin 1 & x \neq 0 \\ 0 & x = 0\end{cases}$$ is differentiable at $x=0$ and find $f'(0)$.

Answer 1

By definition, $$ 0\le |f'(0)|=\lim_{x\to 0} \left|\frac{f(x)-f(0)}{x}\right|\le \lim_{x\to 0}\left|\frac{x^2-0}{x}\right|=\left|\lim_{x\to 0}x\right|=0 $$ (Using the fact that $0\le|f(0)|\le 0^2=0$ so $f(0)=0$)

Answer 2

HINT

For the second one by definition we need to consider

$$f'(0)=\lim_{x\to 0}\frac{x^2\sin 1-f(0)}{x-0}$$

Answer 3

Alternatively you can say that $f= \sin(1)g$ where

$$g(x)=\begin{cases}x^2 & x \neq 0 \\ 0 & x = 0\end{cases}\equiv x^2$$

and the product of two differentiable functions is differentiable (hence the product of a differentiable function by a constant is)