The n-th derivative of $x \cdot \cos(2x)$

Question

I have a function $f(x) = x \cdot \cos(2x)$. I have to find the $n$-th derivative $f^{(n)}(x)$.

I know I have to use Leibnitz's formula, but how to get a general formula for this $f(x)$?

Answer 1

$$\dfrac{d^n(x\cos2x)}{dx^n }=\binom n0x\dfrac{d^n(\cos2x)}{dx^n }+\binom n0\dfrac{dx}{dx}\dfrac{d^{n-1}(\cos2x)}{dx^{n-1} }+0$$

as $\dfrac{d^r(x)}{dx^r}=0$ for $r\ge2$

Method $\#1:$

Now like $100$-th derivative of the function $f(x)=e^{x}\cos(x)$,

$\cos ax$ is the real part of $e^{iax}$

$$\dfrac{d^m(e^{iax})}{dx^m}=(ia)^me^{iax}=a^m(\cos ax+i\sin ax)\left(\cos\dfrac\pi2+i\sin\dfrac\pi2\right)^m$$

Use De Moivre's Theorem Related - Complex number to find the real part

Method $\#2:$

$\dfrac{d(\cos2x)}{dx}=-\sin2x=\cos\left(2x+\dfrac\pi2\right)$

$\dfrac{d^2(\cos2x)}{dx^2}=-\cos2x=\cos\left(2x+2\cdot\dfrac\pi2\right)$

Observe that $\dfrac{d^m(\cos2x)}{dx^m}$ has a period of $4$

Answer 2

Hint:

$\dfrac{d^n(xe^{i2x})}{dx^n}=x(2i)^ne^{i2x}+n(2i)^{n-1}e^{i2x}$

Compare the real parts.