How does one derive the Fourier transform of the Ramp function? One approach could have been to see that the Ramp function ( http://mathworld.wolfram.com/RampFunction.html ) is the convolution of $2$ Heavisides (at $0$). Hence its Fourier transform should have been the product of the Fourier transforms of Heavisides. The Fourier transform of the Heaviside (http://mathworld.wolfram.com/HeavisideStepFunction.html) is, $\frac{1}{2} [\delta(t) - \frac{1}{\pi t}  ] $. But its not clear to me as to how its square is the Fourier transform of the Ramp at $0$ which is $\frac{i}{4\pi} \delta'(t) - \frac{1}{4\pi^2 t^2} $


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*I would otherwise like to see a reference (or if someone can type in!) which derives the Fourier transform of the ramp function from scratch! 


"Frequency derivative" is a property of Fourier transform which is:
$$\mathcal{F}\{x(f(x)\}=j\frac{d}{d\omega}F(\omega)$$
Plug $f(x)=u(x)$ (i.e. heaviside function) whose FT is $F(\omega)=\pi\delta(\omega)-\frac{j}{\omega}$. 
Since $\text{ramp}(x)=xu(x)$ we get
$$\mathcal{F}\{\text{ramp}(x)\}=j\frac{d}{d\omega}\left(\pi\delta(\omega)-\frac{j}{\omega}\right)=j\pi\delta'(\omega)-\frac{1}{\omega^2}$$
If you want to represent it versus $f$, since $\omega=2\pi f$ it becomes
$$\mathcal{F}\{\text{ramp}(x)\}=(j\pi)\frac{1}{(2\pi)^2}\delta'(f)-\frac{1}{4\pi^2f^2}=\frac{j}{4\pi}\delta'(f)-\frac{1}{4\pi^2f^2}$$
