Answer
See explanations.
Work Step by Step
The result from the previous Exercise states that if $f(x)$ is differentiable at $x_0$ with $f(x_0)=0$ and $g(x)$ is continuous at $x_0$, then their product $h(x)=f(x)g(x)$ is differentiable at $x_0$
a. Let $h(x)=|x|\ sin(x)$ and assume $f(x)=sin(x)$ and $g(x)=|x|$. We have $f(x)$ is differentiable at $x=0$ ($f'(x)=cos(x)$) with $f(0)=sin(0)=0$ and $g(x)$ is continuous at $x=0$ ($\lim_{x\to0}g(x)=g(0)$),;hus their product $h(x)=f(x)g(x)$ is differentiable at $x=0$.
b. Let $h(x)=x^{2/3} sin(x)$ and assume $f(x)=sin(x)$ and $g(x)=x^{2/3}$. We have $f(x)$ is differentiable at $x=0$ ($f'(x)=cos(x)$) with $f(0)=sin(0)=0$ and $g(x)$ is continuous at $x=0$ ($\lim_{x\to0}g(x)=g(0)$); thus their product $h(x)=f(x)g(x)$ is differentiable at $x=0$
c. Let $h(x)=\sqrt[3] x\ (1-cos(x))$ and assume $f(x)=(1-cos(x))$ and $g(x)=\sqrt[3] x$. We have $f(x)$ is differentiable at $x=0$ ($f'(x)=sin(x)$) with $f(0)=1-cos(0)=0$ and $g(x)$ is continuous at $x=0$ ($\lim_{x\to0}g(x)=g(0)$); thus their product $h(x)=f(x)g(x)$ is differentiable at $x=0$
d. Given the function $h(x)=\begin{cases} x^2\ sin\frac{1}{x},\ x\ne0\\0, x=0 \end{cases}$, we have $\lim_{x\to0}h(x)=\lim_{x\to0}x^2\ sin\frac{1}{x}=0$ ($|sin\frac{1}{x}|\leq1$); thus we need to show that the function $k(x)=x^2\ sin\frac{1}{x},$ is differentiable at $x=0$. Similarly, assume $f(x)=x$ and $g(x)=x\ sin\frac{1}{x}$. We have $f(x)$ is differentiable at $x=0$ ($f'(x)=1$) with $f(0)=0$ and $g(x)$ is continuous at $x=0$ ($\lim_{x\to0}g(x)=g(0)$); thus their product $h(x)=f(x)g(x)=x^2\ sin\frac{1}{x}$ is differentiable at $x=0$
