The question arises from a problem in Basic Analysis by Lebl.
Suppose $f : I → \mathbb{R}$ is a bounded function and $g : I → \mathbb{R}$ is a function differentiable at $c ∈ I$ and $g(c) = g′ (c) = 0.$ Show that $h(x) := f (x)g(x)$ is differentiable at c. Hint: Note that you cannot apply the product rule.
Using the definition of the derivative I calculated this value to be zero, however I am not sure what effect the boundedness of $f$ has on the situation.
If $f$ is not bounded, the statement may not be true. For example, consider the functions $f$ and $g$ on $I = (-1..1)$ given by $f(x) = 1/x^2$ outside of zero and $f(0) = 0$ an $g(x) = x^2$. Then $g(0) = g'(0) = 0$.
However, the product $h = fg$ is given by $h(x) = 1$ outside of zero and $h(0) = 0$. That is not even continuous at zero, let alone differentiable …
If you didn’t use that $f$ is bounded in your argument, show it and someone might spot a gap or an error.
Since $g(c)=0$ we have $\frac{h(x)-h(c)}{x-c}=\frac{f(x)g(x)-f(x)g(c)}{x-c}=f(x)\frac{g(x)-g(c)}{x-c}$
Since $ \frac{g(x)-g(c)}{x-c} \to g'(c)=0$ for $x \to c$ and $f$ is bounded, we can conclude that
$\frac{h(x)-h(c)}{x-c} \to 0$ for $x \to c$ .
