= Solution
By <differentiability implies continuity>, $f(x+h)\to f(x)$. Add and subtract $f(x+h)g(x)$ in the difference quotient:
$$
\frac{f(x+h)g(x+h)-f(x)g(x)}h
=f(x+h)\frac{g(x+h)-g(x)}h
+g(x)\frac{f(x+h)-f(x)}h.
$$
The two quotient limits exist by the given <differentiability>. Taking the limit and using <continuity> yields the <product rule>:
$$
\boxed{(fg)'(x)=f(x)g'(x)+g(x)f'(x).}
$$
Thus the product is <differentiable> at the specified point; differentiability elsewhere is not needed.
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