<s>[INST] $n$-derivative of function $f(x)=e^x \sin x$ at $x=0$ I have function $f(x) = e^x \sin{x}$ and must found $f^{(n)}(0)$
$f'(x) = e^x(\sin{x} + \cos{x}) $
$f''(x) = 2 e^x \cos{x}$
$f'''(x) = 2 e^x (\cos{x} - \sin{x})$
$f''''(x) = -4 e^x \sin{x}$
$f'''''(x) = -4 e^x (\sin{x} + \cos{x})$
I think $f^{(n)}(0) = \alpha (-1)^n x^{2n+1}$ but I can't find $\alpha$
 [/INST] Let us try to discover a general formula. I see two possible approaches.
The first one is to use induction: assume that $f^{(n)} (x) = \Bbb e ^x (a_n \cos x + b_n \sin x)$. We have $a_0 = 0$ and $b_0 = 1$. Then
$$f^{(n+1)} = \Bbb e ^x (a_n \cos x + b_n \sin x) + \Bbb e ^x (- a_n \sin x + b_n \cos x) = \Bbb e ^x [(a_n + b_n) \cos x + (b_n - a_n) \sin x ] ,$$
which gives us a recurrence relation that can be written in matrix form as
$$\begin{pmatrix} a_{n+1} \\ b_{n+1} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} a_n \\ b_n \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} ^2 \begin{pmatrix} a_{n-1} \\ b_{n-1} \end{pmatrix} = \dots = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} ^{n+1} \begin{pmatrix} a_0 \\ b_0 \end{pmatrix} .$$
If $I$ is the identity matrix, prove by induction (it is easy) that
$$\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} ^{4n} = (-4)^n I .$$
It follows that
$$\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} ^{4n+1} = (-4)^n \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} , \\
\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} ^{4n+2} = (-4)^n \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}^2 = (-4)^n \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} , \\
\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} ^{4n+3} = (-4)^n \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}^3 = (-4)^n \begin{pmatrix} -2 & 2 \\ -2 & -2 \end{pmatrix} ,$$
whence it follows that
$$\begin{pmatrix} a_{4n} \\ b_{4n} \end{pmatrix} = (-4)^n \begin{pmatrix} 0 \\ 1 \end{pmatrix} , \\
\begin{pmatrix} a_{4n+1} \\ b_{4n+1} \end{pmatrix} = (-4)^n \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = (-4)^n \begin{pmatrix} 1 \\ 1 \end{pmatrix} , \\
\begin{pmatrix} a_{4n+2} \\ b_{4n+2} \end{pmatrix} = (-4)^n \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = (-4)^n \begin{pmatrix} 2 \\ 0 \end{pmatrix} , \\
\begin{pmatrix} a_{4n+3} \\ b_{4n+3} \end{pmatrix} = (-4)^n \begin{pmatrix} -2 & 2 \\ -2 & -2 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = (-4)^n \begin{pmatrix} 2 \\ -2 \end{pmatrix} , \\$$
which gives the result
$$f^{(4n)} (x) = \Bbb e^x (-4)^n \sin x , \\
f^{(4n+1)} (x) = \Bbb e^x (-4)^n (\cos x + \sin x), \\
f^{(4n+2)} (x) = \Bbb e^x (-4)^n 2 \cos x , \\
f^{(4n+3)} (x) = \Bbb e^x (-4)^n (2 \cos x - 2 \sin x) ,$$
whence one finally gets
$$f^{(4n)} (0) = 0 , \\
f^{(4n+1)} (0) = (-4)^n , \\
f^{(4n+2)} (0) = 2 (-4)^n , \\
f^{(4n+3)} (0) = 2 (-4)^n .$$

An alternative approach would be to use Leibniz's general formula
$$(fg)^{(n)} = \sum _{k=0} ^n \binom n k f^{(n-k)} g^{(k)}$$
whence it follows that
$$\tag{*} (\Bbb e ^x \sin x) ^{(n)} (0) = \sum _{k=0} ^n \binom n k \sin^{(k)} (0) .$$
It is easy now (again, induction) to see that
$$\sin^{(4n)} = \sin, \quad \sin^{(4n+1)} = \sin' = \cos, \quad \sin^{(4n+2)} = \cos' = -\sin, \quad \sin^{(4n+3)} = -\sin' = -\cos ,$$
meaning that
$$\sin^{(4n)} (0) = 0, \quad \sin^{(4n+1)} (0) = 1, \quad \sin^{(4n+2)} (0) = 0, \quad \sin^{(4n+3)} (0) = -1 .$$
Plugging these in (*) will give you a result, but you would have to work on it a little bit in order to reach the same form obtained through the first method.
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