Answer
a. $f'(0)=0$, see explanations.
b. $f'(0)=0$, see explanations.
Work Step by Step
a. Step 1. To show that $f$ is differentiable at $x=0$, we need to show that the limit $f'(0)=\lim_{x\to0}f'(x)=\lim_{x\to0}\frac{f(x)-f(0)}{x}$ exists.
Step 2. Given $|f(x)|\leq x^2$ for $-1\leq x\leq1$, we have $-x^2\leq f(x)\leq x^2$. Use the Sandwich Theorem, as $\lim_{x\to0}x^2=0$, we have $f(0)=\lim_{x\to0}f(x)=0$
Step 3. The limit in step 1 will be sandwiched as $-x\leq |\frac{f(x)}{x}|\leq x$ and as $\lim_{x\to0}x=0$, we get $f'(0)=\lim_{x\to0}\frac{f(x)}{x}=0$
Step 4. We conclude that $f$ is differentiable at $x=0$ and $f'(0)=0$
b. Step 1. Given $f(x)=x^2sin\frac{1}{x}$ for $x\ne0$, as $|sin\frac{1}{x}|\leq1$, we have $|f(x)|=x^2|sin\frac{1}{x}|\leq x^2$
Step 2. Given $f(x)=0$ at $x=0$, we know that function $f(x)$ is defined at this point.
Step 3. Using the results from part-a, we can conclude that $f$ is differentiable at $x=0$ and $f'(0)=0$ for this case.
