What will be the $n$th derivative of $y = e^{ax} \sin(ax+b)$? $$y = e^{ax} \sin(ax+b)$$
I tried finding a pattern but no luck till the 7th derivative, and I have seen some people solving it with Euler's method but I doubt my familiarity with it. It'd be great if someone could shed some light on a more straight-forward, general and intuitive solution to the problem.
Where a and b are constants.
 A: In a recursive scheme for $y^{(n)}$ we shall have to make room for the factor $e^{ax}$ and for $\sin(ax+b)$ as well for $\cos(ax+b)$ terms. We therefore write
$$y^{(n)}(x)=e^{ax}\bigl(A_n\cos(ax+b)+B_n\sin(ax+b)\bigr)\qquad(n\geq0)\tag{1}$$
with $A_0=0$, $B_0=1$. Differentiating $(1)$ leads to
$$\eqalign{A_{n+1}&=a(A_n+B_n)\cr B_{n+1}&=a(-A_n+B_n)\ .\cr}$$
These equations can be condensed to
$$A_{n+1}+i B_{n+1}=a(1-i)(A_n+iB_n)\ ,$$
so that
$$A_n+iB_n=a^n (1-i)^n\>i\qquad(n\geq0)\ .$$Now $1-i=\sqrt{2}e^{-i\pi/4}$, so that 
$$i(1-i)^n=2^{n/2}e^{(2-n)i\pi/4}\ .$$
This then implies
$$A_n=a^n\>2^{n/2}\cos{(n-2)\pi\over4},\qquad B_n=-a^n\>2^{n/2}\sin{(n-2)\pi\over 4}\qquad(n\geq0)\ .$$
A: Let us denote by $f^{(n)}$ the $n$th derivative of $f$. One has the following:
General Leibniz  Rule: For any $n\in\mathbb{N}$ and $f,g$ two functions $n$ time differentiable, $fg$ is $n$ times differentiable and
$$(fg)^{(n)}=\sum_{k=0}^n\binom{n}{k}f^{(k)}g^{(n-k)}.$$
Now all you need is to find the $n$th derivative of $e^{ax}$ and $\sin(ax+b)$.
A: HINT
One formula to memorise:
$$D^n[e^{ax}.\sin(bx+c)]=(a^2+b^2)^{\frac{n}{2}} e^{ax}\sin(bx+c+n \tan^{-1}\frac{b}{a})$$
How to get that?
$$D^1=e^{ax}[a\sin(bx+c) + b\sin(bx+c+\pi/2)]$$
Let $a=r \cos\alpha, b=r\sin\alpha$ and $r=\sqrt{a^2+b^2}$
Now put it in the equation above:
$$D^1=e^{ax} r\sin(bx+c+\alpha)$$
$$D^n=e^{ax}.r^n.\sin(bx+c+n\alpha)$$
