Product rule

If $f$ and $g$ are differentiable at $a$, then

For a continuous function $f:\mathbb{R}\to \mathbb{R}$, $g(t)=t^{2}f(t)$ is differentiable at zero with $g^{\prime }(0)=0$, because $[g(h)-g(0)]/h=hf(h)\to 0$. At any nonzero point $x$, $g$ is differentiable exactly when $f$ is: one direction uses the product rule, and the converse applies it to $f(t)=t^{-2}g(t)$ near $x$.

The higher-order product rule is

For $m$ functions with sufficiently many derivatives, the $n$th derivative of their product distributes the $n$ differentiations among all factors. If factor $j$ is differentiated $k_j$ times, its term has the multinomial coefficient $n!/(k_1!\cdots k_m!)$, where the nonnegative integers $k_j$ sum to $n$. Induction on the number of factors proves this from the two-factor Leibniz rule: splitting off the final factor gives a binomial coefficient, which multiplies the previous multinomial coefficient to give the one displayed above. Equivalently, count the ways to choose which of the $n$ successive applications of the product rule differentiate each factor.

At zero, $f(x)=x$ and $g(x)=1+|x|$ give a differentiable productwith derivative one, although $g$ is not differentiable.