Answer: For all positive integers \(n\),

\(y^{(n)}=\mathrm{e}^{x}\left(\binom{n}{0}u+\binom{n}{1}u^{(1)}+\binom{n}{2}u^{(2)}+\cdots+\binom{n}{n}u^{(n)}\right)\).

We prove the result by induction on \(n\).

Base case: \(n=1\).

Since \(y=\mathrm{e}^{x}u\), by the product rule

\(y^{(1)}=\dfrac{\mathrm{d}}{\mathrm{d}x}(\mathrm{e}^{x}u)=\mathrm{e}^{x}u+\mathrm{e}^{x}u^{(1)}=\mathrm{e}^{x}\left(\binom{1}{0}u+\binom{1}{1}u^{(1)}\right)\).

So the formula is true when \(n=1\).

Inductive step.

Assume that for some positive integer \(k\),

\(y^{(k)}=\mathrm{e}^{x}\left(\binom{k}{0}u+\binom{k}{1}u^{(1)}+\cdots+\binom{k}{r}u^{(r)}+\cdots+\binom{k}{k}u^{(k)}\right).\)

Differentiate both sides:

\(y^{(k+1)}=\mathrm{e}^{x}\left(\binom{k}{0}u+\binom{k}{1}u^{(1)}+\cdots+\binom{k}{k}u^{(k)}\right)+\mathrm{e}^{x}\left(\binom{k}{0}u^{(1)}+\binom{k}{1}u^{(2)}+\cdots+\binom{k}{k}u^{(k+1)}\right).\)

Now combine like terms. The coefficient of \(u^{(r)}\) for \(1\le r\le k\) is

\(\binom{k}{r}+\binom{k}{r-1}=\binom{k+1}{r}\).

The first term stays as \(\binom{k+1}{0}u\), and the final term is \(\binom{k+1}{k+1}u^{(k+1)}\). Hence

\(y^{(k+1)}=\mathrm{e}^{x}\left(\binom{k+1}{0}u+\binom{k+1}{1}u^{(1)}+\cdots+\binom{k+1}{k+1}u^{(k+1)}\right).\)

This is exactly the required formula with \(n=k+1\).

Therefore, if the result holds for \(n=k\), it also holds for \(n=k+1\). Since it is true for \(n=1\), it follows by mathematical induction that for all positive integers \(n\),

\(y^{(n)}=\mathrm{e}^{x}\left(\binom{n}{0}u+\binom{n}{1}u^{(1)}+\binom{n}{2}u^{(2)}+\cdots+\binom{n}{n}u^{(n)}\right).\)