Finding the $n$-th derivative of $f(x)=e^{x}\sin(x)$ I am trying to find the general form for the $n$-th derivative of $f(x)=e^{x}\sin(x)$.
I have calculated the derivatives up to $5$, but I am having trouble coming up with a general rule. Here is my work so far: 
$$\begin{align}
f^{(1)}(x)&=e^{x}\cos(x) + e^{x} \sin(x)\\
f^{(2)}(x)&= 2e^{x}\cos(x)= e^{x}(\sin(x)+ \cos(x)) + e^{x}(\cos(x)- \sin(x))\\
f^{(3)}(x)&=2e^{x}\cos(x) -2e^{x} \sin(x)\\
f^{(4)}(x)&=-4e^{x}\sin(x)\\
f^{(5)}(x)&=-4e^{x}\cos(x) -4e^{x} \sin(x)\\
\end{align}$$
I am having trouble spotting the pattern to derive the formula for the nth derivative of this function. Any help would be greatly appreciated.

The pattern is already visible —
observe that $f^{4}(x)=-4f(x)$, thus
$$f^{(n)}(x)=(-4)^{[n/4]}f^{(n \bmod 4)}(x)$$
