$n$th derivative of $e^{ax}\sin(bx+c)$ 

  
*
  
*How can we substitute $r \cos \alpha$ and $r \sin \alpha$ for $a$ and $b$?
  
*How, on successive differentiation, is there another $r$ multiplied? 
  

 A: $$y=~e^{ax}\sin(bx+c)~$$
Differentiating with respect to $~x~$, we have 
$$y_1=\frac{dy}{dx}=~a~e^{ax}\sin(bx+c)~+~b~e^{ax}\cos(bx+c)~$$
$$\implies y_1=~e^{ax}~\{~a~\sin(bx+c)~+~b~\cos(bx+c)\}~\tag1$$
For computation of higher-order derivatives it is convenient to express the constants $~a~$ and $~b~$ in terms of the constants $~r~$ and $~\alpha~$ defined by
$$a=r\cos\alpha,~~~~~~~~b=r\sin\alpha$$so that $$r=\sqrt{a^2+b^2},\qquad\text{and}\qquad \alpha=\tan^{-1}\left(\frac{b}{a}\right)$$
Then $~(1)~$ implies, 
$$y_1=~e^{ax}~\{~r\cos\alpha\cdot~\sin(bx+c)~+~r\sin\alpha\cdot~\cos(bx+c)\}$$
$$\implies y_1=~r~~e^{ax}\{\cos\alpha\cdot~\sin(bx+c)~+~\sin\alpha\cdot~\cos(bx+c)\}$$
$$\implies y_1=~r~e^{ax}\sin(bx+c+\alpha)$$
Therefore $$y_2=\frac{d^2y}{dx^2}=~r~\{~a~e^{ax}\sin(bx+c+\alpha)~+~b~e^{ax}\cos(bx+c+\alpha)\}$$
$$\implies y_2=~r~\{~r\cos\alpha\cdot~e^{ax}\sin(bx+c+\alpha)~+~r\sin\alpha\cdot~e^{ax}\cos(bx+c+\alpha)\}$$
$$\implies y_2=~r^2~e^{ax}~\{\cos\alpha\cdot\sin(bx+c+\alpha)~+~r\sin\alpha\cdot\cos(bx+c+\alpha)\}$$
$$\implies y_2=~r^2~e^{ax}~\sin(bx+c+2\alpha)$$
Proceeding like this, we obtain $$y_n=\frac{d^ny}{dx^n}=~r^n~e^{ax}~\sin(bx+c+n\alpha)$$
Putting the values of $~r,~\alpha~$ in the latter equation we have
$$y_n=~e^{ax}~(a^2+b^2)^{\frac{n}{2}}~\sin\left(bx+c+n~\tan^{-1}\frac{b}{a}\right)$$
A: Hint
An easier solution is $$f(x)=e^{ax}\sin (bx+c)=\Im\{ e^{ax+jbx+jc}\}$$therefore $${d^n f(x)\over dx^n}=\Im\{{d^n\over dx^n} e^{(a+jb)x+jc}\}=\Im\{e^{jc}{d^n\over dx^n}e^{(a+jb)x}\}=\Im \{e^{jc}(a+jb)^ne^{(a+jb)x}\}$$
A: First, note that the title in the excerpt is incorrect. It should be "$n$th derivative of $e^{ax}\sin(bx+c)$".
(1) Here we introducing new quantities $r$ and $\alpha$, so we may define them however we want. Essentially, this amounts to writing the pair $(a, b)$ in polar coordinates, as $(r, \alpha)_{\textrm{polar}}$.
More explicitly, any such $r$ satisfies $$a^2 + b^2 = (r \cos \alpha)^2 + (r \sin \alpha)^2 = r^2 (\cos^2 \alpha + \sin^2 \alpha) = r^2 ,$$
so, $$r = \pm \sqrt{a^2 + b^2}.$$
Now the length of $$\left(\frac{a}{r}, \frac{b}{r}\right)$$ is
$$\sqrt{\left(\frac{a}{r}\right)^2 + \left(\frac{b}{r}\right)^2} = \sqrt{\frac{a^2 + b^2}{r^2}} = \sqrt{\frac{r^2}{r^2}} = 1 ,$$
and so it lies on the unit circle. In particular, since $\theta \mapsto (\cos \theta, \sin \theta)$ parameterizes all of unit circle, there is some $\alpha$ such that $$\frac{a}{r} = \cos \alpha, \qquad \frac{b}{r} = \sin \alpha,$$
and rearranging gives the equations in the text, namely,
$$a = r \cos \alpha, \qquad b = r \sin \alpha .$$
Since $(r \cos (\alpha + \pi), r \sin(\alpha + \pi)) = (-r \cos \alpha, -r \sin \alpha)$, by possibly adding $\pi$ to $\alpha$ we may as well assume that $r \geq 0$, that is that $r = \sqrt{a^2 + b^2}$.
(2) The computation for $y_1$ gives that
$$y_1 = r e^{\alpha x} \sin (b x + c + \alpha) .$$
If we want to compute the next derivative, we have
$$y_2 = (r e^{\alpha x} \sin (b x + c + \alpha))' = r (e^{\alpha x} \sin (b x + c + \alpha))$$
But if we replace $c$ with $c + \alpha$ in our rule, it tells us that $(e^{\alpha x} \sin (b x + c + \alpha) = r e^{\alpha x} \sin (b x + c + 2 \alpha)$, and substituting back in the previous display equation gives 
$$y_2 = r (r e^{\alpha x} \sin (b x + (c + \alpha) + \alpha)) = r^2 e^{\alpha x} \sin (b x + c + 2 \alpha)$$
as claimed.
