Answer: For every positive integer \(n\),
\(\displaystyle \frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(\mathrm{e}^{x}\sin x)=(\sqrt{2})^{n}\mathrm{e}^{x}\sin\left(x+\frac{n\pi}{4}\right).\)
Let \(H_n\) be the statement
\(\displaystyle \frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(\mathrm{e}^{x}\sin x)=(\sqrt{2})^{n}\mathrm{e}^{x}\sin\left(x+\frac{n\pi}{4}\right).\)
We prove \(H_n\) by induction.
First, when \(n=1\),
\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(\mathrm{e}^{x}\sin x)=\mathrm{e}^{x}\sin x+\mathrm{e}^{x}\cos x.\)
Now use \(\sin\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x\). Hence
\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(\mathrm{e}^{x}\sin x)=\sqrt{2}\,\mathrm{e}^{x}\sin\left(x+\frac{\pi}{4}\right),\)
so \(H_1\) is true.
Now assume that \(H_k\) is true for some positive integer \(k\), so
\(\displaystyle \frac{\mathrm{d}^{k}}{\mathrm{d}x^{k}}(\mathrm{e}^{x}\sin x)=(\sqrt{2})^{k}\mathrm{e}^{x}\sin\left(x+\frac{k\pi}{4}\right).\)
Differentiating both sides gives
\(\displaystyle \frac{\mathrm{d}^{k+1}}{\mathrm{d}x^{k+1}}(\mathrm{e}^{x}\sin x)=(\sqrt{2})^{k}\frac{\mathrm{d}}{\mathrm{d}x}\left[\mathrm{e}^{x}\sin\left(x+\frac{k\pi}{4}\right)\right].\)
Using the product rule,
\(\displaystyle \frac{\mathrm{d}^{k+1}}{\mathrm{d}x^{k+1}}(\mathrm{e}^{x}\sin x)=(\sqrt{2})^{k}\mathrm{e}^{x}\left[\sin\left(x+\frac{k\pi}{4}\right)+\cos\left(x+\frac{k\pi}{4}\right)\right].\)
Factor out \(\sqrt{2}\):
\(\displaystyle =(\sqrt{2})^{k+1}\mathrm{e}^{x}\left[\frac{1}{\sqrt{2}}\sin\left(x+\frac{k\pi}{4}\right)+\frac{1}{\sqrt{2}}\cos\left(x+\frac{k\pi}{4}\right)\right].\)
Now use the identity \(\sin(u+\frac{\pi}{4})=\frac{1}{\sqrt{2}}\sin u+\frac{1}{\sqrt{2}}\cos u\), with \(u=x+\frac{k\pi}{4}\). This gives
\(\displaystyle \frac{\mathrm{d}^{k+1}}{\mathrm{d}x^{k+1}}(\mathrm{e}^{x}\sin x)=(\sqrt{2})^{k+1}\mathrm{e}^{x}\sin\left(x+\frac{(k+1)\pi}{4}\right).\)
So \(H_{k+1}\) is true whenever \(H_k\) is true.
Therefore, by mathematical induction, \(H_n\) is true for every positive integer \(n\).
