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.

Write $$f(x) = e^x \sin x = \frac{1}{2i} e^x (e^{ix} - e^{-ix}) = \frac{1}{2i} (e^{(1+i)x} - e^{(1-i)x}).$$  Then we immediately have $$f^{(n)}(x) = \frac{d^n}{dx^n}\left[e^x \sin x\right] = \frac{1}{2i} \left((1+i)^n e^{(1+i)x} - (1-i)^n e^{(1-i)x}\right).$$  All that remains is to rewrite this in terms of real valued functions.  We note $1 + i = \sqrt{2} e^{\pi i/4}$ and $1-i = \sqrt{2} e^{-\pi i /4}$, so $$f^{(n)}(x) = \frac{e^x}{2i} 2^{n/2} \left( e^{ix + \pi i n/4} - e^{-(ix + \pi i n/4)} \right) = e^x 2^{n/2} \sin \left(x + \frac{n\pi}{4}\right).$$
