# Time Differenciation property of the Fourier Transform

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Given that $F(\omega)=\mathcal{F}[f(t)]$, then $$\mathcal{F}\left[f^{\prime}(t)\right]=j \omega F(\omega)$$ In other words, the transform of the derivative of $f(t)$ is obtained by multiplying the transform of $f(t)$ by $j \omega$. By definition,
$$f(t)=\mathcal{F}^{-1}[F(\omega)]=\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{j \omega t} d \omega$$
Taking the derivative of both sides with respect to $t$ gives
$$f^{\prime}(t)=\frac{j \omega}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{j \omega t} d \omega=j \omega \mathcal{F}^{-1}[F(\omega)] \tag{1}$$
or $$\mathcal{F}\left[f^{\prime}(t)\right]=j \omega F(\omega)$$ Repeated applications of Eq. (1) give $$\mathcal{F}\left[f^{(n)}(t)\right]=(j \omega)^{n} F(\omega)$$ For example, if $f(t)=e^{-a t}$, then $$f^{\prime}(t)=-a e^{-a t}=-a f(t)$$ Taking the Fourier transforms of the first and last terms, we obtain
$$j \omega F(\omega)=-a F(\omega) \quad \Longrightarrow \quad F(\omega)=\frac{1}{a+j \omega}$$

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