The Fourier transform of this $e^{-2\pi i f_0 t}$ is:
$$\delta(f-f_0)$$
One can just imagine this as a vertical line in Fourier Space picking out the specific frequency of that exponential.
The Fourier Transform of that complex exponential times $t$ is:
$$\frac{i}{2\pi}\delta'(f-f_0) =\frac{i}{2\pi}\delta(f-f_0)\frac{d}{df} $$
Clearly this is an operator, but I would like to imagine how this would look when plotted. Any ideas?
Distributions are limits of sequences of functions, the limit being taken in the sense of distributions
$$\delta(x)=\lim_{n\to \infty} n e^{-\pi n^2 x^2},\qquad\delta'(x)=\lim_{n\to \infty} (n e^{-\pi n^2 x^2})'=\lim_{n\to \infty} - \pi n^3 2x e^{-\pi n^2 x^2}$$ Alternatively you can visualize it as two very close $\delta$ $$\delta'(x) = \lim_{h\to 0} \frac{\delta(x+h)-\delta(x-h)}{2h}$$