Justify if $fg$ is differentiable at point $x_0=0$ I had these two statements in some exam recently and I still wonder how to tackle them to decide which one is correct.


*

*$f$ is continuous at $0$, $g$ is differentiable at $0$, $g(0)=0$, $fg$ is differentiable at $0$.

*$f$ is continuous at $0$, $g$ is differentiable at $0$, $f(0)=0$, $fg$ is differentiable at $0$.
I know that if $g$ is differentiable at a certain point $x_0$, then $g$ must also be continuous at $x_0$. Also, $f$ is differentiable at $x_0$ iff $\lim \limits_{x \to x_0 } \frac{f(x) - f(x_0)}{x - x_0}$ exists and that value is denoted by $f'(x_0)$. But I'm confused about the importance of $g(0)=0$ and $f(0)=0$.
 A: 
I'm confused about the importance of $g(0)=0$ and $f(0)=0$.

Here's the idea of why that's important. Suppose we try applying the product rule. Then
$$ (fg)'(0) = f(0)g'(0) + f'(0)g(0) $$
Now we don't know that $f'(0)$ exists so if $g(0) = 0$ then we "don't have to worry about the last term." That is, if $g(0) = 0$ we might expect
$$ (fg)'(0) = f(0)g'(0). $$
A: Hint:


*

*define the function $h$ as $h = fg$.

*Write down, by definition, what it means for $h$ to be differentiable at $0$.

*Write down, by definition, what it means for $g$ to be differentiable at $0$.

*Compare the two expressions from $2$ and $3$.


What did you get as the result of step $2$? What did you get as the result of step $3$? How do the two compare?
A: I will give you a "proof" that is a bit harder than you might hope, but I invite you to think about it.
Since $g$ is differentiable at zero, we can write
$$
g(x)=g(0)+g'(0)x + o(x) \tag{1}
$$
as $x \to 0$. Here $\lim_{x \to 0} \frac{o(x)}{x}=0$. Now,
$$
f(x)g(x) = f(x) \left( g(0)+g'(0)x+o(x) \right) = f(x)g(0) + g'(0)f(x)x + f(x)o(x). \tag{2}
$$
We must decide whether the right-hand side of (2) is an affine function plus $o(x)$ as $x \to 0$. But
$$
\lim_{x \to 0} \frac{f(x)g(x)}{x} = \lim_{x \to 0} \frac{f(x)g(0) + g'(0)f(x)x + f(x)o(x)}{x}.
$$
It follows easily that if $g(0)=0$, then
$$
\lim_{x \to 0} \frac{f(x)g(x)}{x} = g'(0)f(0)
$$
due to the continuity of $f$. The correct answer should be 1.
