The $257^{\text{th}}$ derivative of $e^{-t} \sin t$ How to find the $257^{\text{th}}$ derivative of $e^{-t} \sin t$.
I got the wrong values in the end. Not very sure how to go on after calculating $2^{\large\frac{257}{2}}e^{\large\frac{3i}{4}}$.
 A: \begin{align}
e^{-t}\sin t & = \Im\left[e^{(i - 1)t}\right]\\
\dfrac{d^{257}}{dt^{257}} e^{-t}\sin t & = \Im\left[ (i - 1)^{257}e^{(i-1)t} \right]\\
& = \Im\left[ 2^{257/2} \left(\dfrac{i - 1}{\sqrt 2}\right)^{257}  e^{(i - 1)t}\right]\\
& = \Im\left[ 2^{257/2} \left(- \cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}\right)^{257}e^{it}e^{-t}\right]\\
& = \Im\left[ -2^{257/2} e^{-t} \left(\cos \dfrac{257\pi}{4} - i \sin \dfrac{257\pi}{4}\right) (\cos t + i \sin t) \right]
\end{align}
Multiplying, and extracting the imaginary part:
\begin{align}
\dfrac{d^{257}}{dt^{257}} e^{-t}\sin t & = -2^{128}\sqrt{2}e^{-t}\left(\cos \dfrac{257\pi}{4}\sin t - \sin\dfrac{257\pi}{4}\cos t \right)\\
& = -2^{128}\sqrt{2}e^{-t}\sin \left( t - \dfrac{257\pi}{4} \right)\\
& = -2^{128}\sqrt{2}e^{-t}\sin \left( t - \dfrac{\pi}{4} - 64\pi\right)\\
& = 2^{128}\sqrt{2}e^{-t}\sin \left(\dfrac{\pi}{4} -t\right)\\
& = \boxed{2^{128}e^{-t}(\cos t - \sin t)}
\end{align}
A: From $f(t)=e^{-t}\sin(t)$, differentiate twice to get $f^{''}(t)={-2}\ e^{-t}\cos(t)$. From $g(t)=e^{-t}\cos(t)$, differentiate twice to get $g^{''}(t)={2}\ e^{-t}\sin(t)$ and hence because $f^{''}(t)={-2}\ g(t)$ it follow that  $f^{iv}(t)={-4}\ e^{-t}\sin(t) = -4 f(t)$.
Iterating this 64 times we get 
$$\dfrac{d^{256}}{dt^{256}}f(t)=(-4)^{64}f(t)$$
Differentiating one more time gives the answer:
\begin{align}
\dfrac{d^{257}}{dt^{257}}f(t)&=2^{128}f^{'}(t) \\
&= 2^{128}e^{-t}(\cos(t)-\sin(t)) \\
&= 2^{128}\sqrt{2}e^{-t}\cos(t+{\pi\over 4})\\
&=2^{128}\sqrt{2}e^{-t}\sin (\dfrac{\pi}{4} -t) \ 
\end{align}
The last line follows because $ \sin(\dfrac{\pi}{2} -t) = \cos(t)$.
