I don't get certain of parts of these two questions
1) I'm trying to do the Fourier transform of: $$f(x) = \, xe^{-x^2} $$
In the problem it said to use:
$$F \, (e^{-tx^2}) = {\sqrt{\frac{π}t}e^{{\frac{-λ^2}{4t}}}} $$
The solution shows this for the first line:
$$F \, (xe^{-x^2}) = F \Big ({-{\frac{1}2}[e^{{{-x^2}}}]'} \Big ) $$
I'm confused about this first part of the solution where it said that it was done by linearity. I don't get how -1/2 came from and how it was derived to get prime('). I do understand what was done afterwards.
2) This is the second problem: $$f(x) = \, e^{-0.5|x|} $$
in the solution it said:
let $$f(x) = \, e^{-|x|} $$ Then $$F(e^{-0.5|x|}) = \,F (f(0.5x)) = \frac {1}{0.5} \hat{f}(\frac{λ}{0.5}) = \frac{4}{1+4λ^2}$$
I'm confused about how $$\frac {1}{0.5} \hat{f}(\frac{λ}{0.5})= \frac{4}{1+4λ^2}$$ What was done with the $$\hat{f}$$? but I do understand that $$f(x) = \, e^{-|x|} = \frac{2}{1+λ^2}$$
To make Fourier Transforms easy there are a few simple rules like a rule for scaling, modulation, translation and so on. I will use these rules as we go on. They are fundamental when doing Fourier-transforms and you should be familiar with them. Here on the first page you can find many of them http://www.ibk.ethz.ch/ch/education/identmeth/fourier.pdf
In the first example we are going to use the derivation rule: $$ \mathscr{F}\left(tf(t)\right) = i\frac{d}{d\omega}\mathscr{F}\left(f(t)\right)(\omega) $$ After applying that you get $$ \mathscr{F}\left(xe^{-x^2}\right) = i\frac{d}{d\omega}\mathscr{F}\left(e^{-x^2}\right)(\omega) $$ $\mathscr{F}\left(e^{-x^2}\right)$ can be easily calculated to $\sqrt{\pi}e^{-\omega^2/4}$ and thus Fouriertransform of your problem is $$ -i\frac{1}{2}\omega\sqrt{\pi}e^{-\omega^2/4} $$
In the second problem, you have to use the scaling rule that states $$ \mathscr{F}\left(f(at)\right) = \frac{1}{|{a}|}\mathscr{F}\left(f(t)\right)\left(\frac{\omega}{a}\right) $$ This means that you have to scale by $a = \frac{1}{2}$ $$ \mathscr{F}\left(e^{-\frac{1}{2}|x|}\right) = \frac{1}{\left|{\frac{1}{2}}\right|}\mathscr{F}\left(e^{-|x|}\right)\left(\frac{\omega}{\frac{1}{2}}\right) $$ Again, $\mathscr{F}\left(e^{-|x|}\right)$ can be easily calculated to $\frac{2}{1+\omega^2}$, which in the end gives $$ \mathscr{F}\left(e^{-\frac{1}{2}|x|}\right) = \frac{4}{1+4\omega^2} $$