What will be the $n$th derivative of $y = e^{ax} \sin(ax+b)$?

Question

$$y = e^{ax} \sin(ax+b)$$

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.

Have you tried writing $\sin(ax+b)$ as a sum of complex exponentials? — lulu, Aug 16 '17 at 17:09
Hello, I think you are pointing out to Euler's formula which I am unaware of/not familiar with. — DmRo912, Aug 16 '17 at 17:10
Euler's formula : $\forall t\in\mathbb R,;\sin t = \dfrac{e^{it}-e^{-it}}{2i}$. — paf, Aug 16 '17 at 17:11
See https://math.stackexchange.com/questions/2393139/how-to-find-nth-derivative-of-this-y/2393143#2393143 — lab bhattacharjee, Aug 16 '17 at 17:12
The pattern is of period $8$. If you refuse to deal with complex numbers you'll never discern it. — Christian Blatter, Aug 16 '17 at 18:13
@ChristianBlatter. "The shortest path to truth in $\mathbb R$ is through $\mathbb C$."--- Hadamard. — DanielWainfleet, Aug 16 '17 at 19:11
Indeed the complex approach allows to simplify massively the (rather non practical) approach suggested in the accepted answer and it has been explained in no answer below. But the OP is not interested... — Did, Oct 19 '17 at 19:34

score 3 · Answer 1 · answered Aug 16 '17 at 19:36

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 $$A_n=a^n\>2^{n/2}\cos{(n-2)\pi\over4},\qquad B_n=-a^n\>2^{n/2}\sin{(n-2)\pi\over 4}\qquad(n\geq0)\ .$$

score 1 · Accepted Answer · answered Aug 16 '17 at 17:14

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 $\sin(ax+b)$.

score 1 · Answer 3 · answered Aug 16 '17 at 17:15

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: $$D^1=e^{ax} r\sin(bx+c+\alpha)$$ $$D^n=e^{ax}.r^n.\sin(bx+c+n\alpha)$$

What will be the $n$th derivative of $y = e^{ax} \sin(ax+b)$?

3 Answers3