Use complex exponential to find $n$th derivative of $\cos^2(ax)$

Question

I'm stuck. Any idea as to how I can continue?

$\cos(2ax) + i\sin(2ax) = e^{2iax}= (e^{iax})^2 = (\cos(ax)+i\sin(ax))^2$

Answer 1

We have$$\cos^2(ax)=\left(\frac{e^{iax}+e^{-iax}}2\right)^2=\frac{e^{2iax}+2+e^{-2iax}}4.$$Therefore, the $n$^th derivative od $\cos^2(ax)$ is$$\frac{(2ia)^ne^{2iax}+(-2ia)^ne^{-2iax}}4.$$Can you take it from here?

Answer 2

Hint

$$ \cos(a x) = \frac{1}{2}\left(e^{i ax}+e^{-i a x}\right) $$

we have

$$ \left(\frac{1}{2}\left(e^{i ax}+e^{-i a x}\right)\right)^2 = \frac{1}{2}+\frac{1}{4}\left(e^{2iax}+e^{-2iax}\right) $$

Now deriving $e^{2iax}$...

Answer 3

$$\cos^2ax=\frac{1+\cos 2ax}2$$ so that $$\frac{d^n}{dx^n}\cos^2ax=2^{n-1}a^n\cos(2ax+n\pi/2).$$ You could rewrite this as $$\cos^2ax=\frac{2+e^{2iax}+e^{-2iax}}4$$ so that $$\frac{d^n}{dx^n}\cos^2ax=2^{n-2}i^na^n(e^{2iax}+(-1)^ne^{-2iax})$$ if you really, really want to use complex exponenials.