Solution
= Solution
Differentiate the product in the definition of $H_n$. Using the identity in (b),
$$
\begin{aligned}
H_n'&=(-1)^ne^{x^2}\left(2xu^{(n)}+u^{(n+1)}\right)\\
&=-2n(-1)^ne^{x^2}u^{(n-1)}
=2n(-1)^{n-1}e^{x^2}u^{(n-1)}.
\end{aligned}
$$
The last expression is $2nH_{n-1}$, so
$$
\boxed{H_n'=2nH_{n-1}\qquad(n\geq1)}.
$$
Also $H_0'=0$ separately. This is the <Hermite polynomial derivative recurrence> in the physicists' normalization.