Properties of the Fourier Transform - Time Differentitation

[p75213]

Homework Statement

This is copied from a book:
$$\eqalign{
& {\rm{Time Differentitation}} \cr
& {\rm{Given that: }}F(\omega ) = F\left[ {f(t)} \right] \cr
& F\left[ {f'(t)} \right] = jwF(\omega ) \cr
& {\rm{Proof:}} \cr
& f(t) = {F^{ - 1}}\left[ {F\left( \omega \right)} \right] = {1 \over {2\pi }}\int_{ - \infty }^\infty {F\left( \omega \right){e^{j\omega t}}d\omega } \cr
& {\rm{Taking the derivative of both sides with respect to }}t{\rm{ gives:}} \cr
& {d \over {dt}}f(t) = {{j\omega } \over {2\pi }}\int_{ - \infty }^\infty {F\left( \omega \right){e^{j\omega t}}d\omega } = j\omega {F^{ - 1}}\left[ {F(\omega )} \right]{\rm{ or }}F\left[ {f'(t)} \right] = jwF(\omega ) \cr} $$

Can somebody explain why the jw is outside the integral? I can't see how that happens using Leibniz's integral rule - http://en.wikipedia.org/wiki/Leibniz_integral_rule

Homework Equations

The Attempt at a Solution

 

[marcusl]

The factor of jw should be inside the integral, as you surmise. Then take the forward FT of both sides to get the answer.

 

[p75213]

Thanks marcusl. I see now.