Second derivative of a product of exponential and sine functions

Using the product rule, $(uv)' = u'v + uv'$. For $f(x) = e^x \sin(x)$, we have $f'(x) = (e^x)'\sin(x) + e^x(\sin(x))' = e^x \sin(x) + e^x \cos(x) = e^x(\sin x + \cos x)$.

Now differentiate $f'(x) = e^x(\sin x + \cos x)$ using the product rule again: $f''(x) = (e^x)'(\sin x + \cos x) + e^x(\sin x + \cos x)' = e^x(\sin x + \cos x) + e^x(\cos x - \sin x)$. Expanding: $f''(x) = e^x \sin x + e^x \cos x + e^x \cos x - e^x \sin x$. The terms $e^x \sin x$ and $-e^x \sin x$ cancel, leaving $f''(x) = 2e^x \cos x$.