One of my friends asked this question. Initially, I also thought that it was obvious to take the derivative of only the function but now I am starting to have doubt about it.
Consider $F(K)$ is the Fourier transform of $f(x)$.
$$F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)e^{ikx}\mathrm{d}x$$
so while taking derivative of its first order
$$F'(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f'(x)e^{ikx}dx $$
But why isn't the $e^{ikx}$ also not differentiated since $x$ is present there? Why isn't the chain rule applied to it? I hope this doesn't sound insane.
Thank you.
In the formula $$F(\omega):={1\over\sqrt{2\pi}}\int_{-\infty}^\infty f(x)\>e^{-i\omega x}\>dx$$ the variable $\omega$ acts as a parameter in the integral on the RHS. Differentiating under the integral sign with respect to this parameter (and neglecting questions of convergence etc.) gives $$F'(\omega):={1\over\sqrt{2\pi}}\int_{-\infty}^\infty (-ix) f(x)\>e^{-i\omega x}\>dx\ .$$