Fourier Transform of $f'(x)$

Question

If the Fourier transform of the function $f(x)$ is $F(k)$, find the Fourier transform of function $f'(x) = [f(-x)]^*$

I am unsure if $f'(x)$ in this case is the derivative function of $f(x)$ or just an entire other function.

If $f'(x)$ is the derivative of $f(x)$ how do I approach this question?

Answer 1

By definition, $$\mathcal{F}^{-1}\{F(k)\}=f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}F(k)e^{ikx}dk\tag{1}$$

Assume $\mathcal{F}\{(f(-x))^*\}=G(k)$. Thus

$$(f(-x))^*=\frac{1}{2\pi}\int_{-\infty}^{+\infty}G(k)e^{ikx}dk$$ $$f(-x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}G^*(k)e^{-ikx}dk$$ $$f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}G^*(k)e^{ikx}dk\tag{2}$$

Comparing $(2)$ with $(1)$ we have

$$G(k)=\mathcal{F}\{(f(-x))^*\}=F^*(k)$$