Answer: Let \(H_n\) be the statement that \(\dfrac{\mathrm d^n}{\mathrm d x^n}(e^x\sin x)=2^{n/2}e^x\sin\left(x+\dfrac{n\pi}{4}\right)\).

For \(n=1\),

\(\dfrac{\mathrm d}{\mathrm d x}(e^x\sin x)=e^x\sin x+e^x\cos x\),

and since \(\sin x+\cos x=\sqrt2\sin\left(x+\dfrac{\pi}{4}\right)\), this gives

\(\dfrac{\mathrm d}{\mathrm d x}(e^x\sin x)=2^{1/2}e^x\sin\left(x+\dfrac{\pi}{4}\right)\).

So \(H_1\) is true.

Now assume \(H_k\) is true for some positive integer \(k\), so that

\(\dfrac{\mathrm d^k}{\mathrm d x^k}(e^x\sin x)=2^{k/2}e^x\sin\left(x+\dfrac{k\pi}{4}\right)\).

Differentiate both sides:

\(\dfrac{\mathrm d^{k+1}}{\mathrm d x^{k+1}}(e^x\sin x)=2^{k/2}\left(e^x\sin\left(x+\dfrac{k\pi}{4}\right)+e^x\cos\left(x+\dfrac{k\pi}{4}\right)\right)\).

Factor out \(2^{(k+1)/2}e^x\):

\(\dfrac{\mathrm d^{k+1}}{\mathrm d x^{k+1}}(e^x\sin x)=2^{(k+1)/2}e^x\left(\dfrac{1}{\sqrt2}\sin\left(x+\dfrac{k\pi}{4}\right)+\dfrac{1}{\sqrt2}\cos\left(x+\dfrac{k\pi}{4}\right)\right)\).

Using \(\sin A\cos\dfrac{\pi}{4}+\cos A\sin\dfrac{\pi}{4}=\sin\left(A+\dfrac{\pi}{4}\right)\), this becomes

\(\dfrac{\mathrm d^{k+1}}{\mathrm d x^{k+1}}(e^x\sin x)=2^{(k+1)/2}e^x\sin\left(x+\dfrac{(k+1)\pi}{4}\right)\).

This is exactly \(H_{k+1}\). Therefore \(H_k\Rightarrow H_{k+1}\).

Hence, by mathematical induction,

\(\dfrac{\mathrm d^{n}}{\mathrm d x^{n}}\left(e^x\sin x\right)=2^{n/2}e^x\sin\left(x+\dfrac{n\pi}{4}\right)\)

for all positive integers \(n\).

Let \(H_n\) denote the statement

\(\dfrac{\mathrm d^{n}}{\mathrm d x^{n}}\left(e^x\sin x\right)=2^{n/2}e^x\sin\left(x+\dfrac{n\pi}{4}\right).\)

Base case. For \(n=1\),

\(\dfrac{\mathrm d}{\mathrm d x}(e^x\sin x)=e^x\sin x+e^x\cos x.\)

Now \(\sin x+\cos x=\sqrt2\sin\left(x+\dfrac{\pi}{4}\right)\), so

\(\dfrac{\mathrm d}{\mathrm d x}(e^x\sin x)=\sqrt2\,e^x\sin\left(x+\dfrac{\pi}{4}\right)=2^{1/2}e^x\sin\left(x+\dfrac{\pi}{4}\right).\)

Thus \(H_1\) is true.

Inductive step. Assume \(H_k\) is true for some positive integer \(k\), so

\(\dfrac{\mathrm d^{k}}{\mathrm d x^{k}}\left(e^x\sin x\right)=2^{k/2}e^x\sin\left(x+\dfrac{k\pi}{4}\right).\)

Differentiate both sides with respect to \(x\):

\(\dfrac{\mathrm d^{k+1}}{\mathrm d x^{k+1}}\left(e^x\sin x\right)=2^{k/2}e^x\sin\left(x+\dfrac{k\pi}{4}\right)+2^{k/2}e^x\cos\left(x+\dfrac{k\pi}{4}\right).\)

Factor out \(2^{(k+1)/2}e^x\):

\(\dfrac{\mathrm d^{k+1}}{\mathrm d x^{k+1}}\left(e^x\sin x\right)=2^{(k+1)/2}e^x\left(\dfrac{1}{\sqrt2}\sin\left(x+\dfrac{k\pi}{4}\right)+\dfrac{1}{\sqrt2}\cos\left(x+\dfrac{k\pi}{4}\right)\right).\)

Use the identity \(\sin A\cos\dfrac{\pi}{4}+\cos A\sin\dfrac{\pi}{4}=\sin\left(A+\dfrac{\pi}{4}\right)\), with \(A=x+\dfrac{k\pi}{4}\). Since \(\cos\dfrac{\pi}{4}=\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}\), the bracket becomes

\(\sin\left(x+\dfrac{k\pi}{4}+\dfrac{\pi}{4}\right)=\sin\left(x+\dfrac{(k+1)\pi}{4}\right).\)

Therefore

\(\dfrac{\mathrm d^{k+1}}{\mathrm d x^{k+1}}\left(e^x\sin x\right)=2^{(k+1)/2}e^x\sin\left(x+\dfrac{(k+1)\pi}{4}\right),\)

which is \(H_{k+1}\).

Since \(H_1\) is true and \(H_k\Rightarrow H_{k+1}\), the result follows by mathematical induction for all positive integers \(n\).