Taking the derivative under a principal value integral

I will assume you are trying to show $$\frac{d}{d t} \mathrm{P} \int_{-\infty}^\infty dx \ \frac{\phi(x)}{x-t} = \mathrm{P} \int_{-\infty}^\infty dx \ \frac{\phi(x)-\phi(t)}{(x-t)^2}$$ Using the definition of principal value and the Leibniz integral rule you will find $$\frac{d}{d t} \mathrm{P} \int_{-\infty}^\infty dx \ \frac{\phi(x)}{x-t} = -\lim_{\epsilon\rightarrow 0}\frac{2}{\epsilon} \phi(t) + \mathrm{P} \int_{-\infty}^\infty dx \ \frac{\phi(x)}{(x-t)^2}$$ Then notice that $$\mathrm{P} \int_{-\infty}^\infty dx \ \frac{1}{(x-t)^2} = \frac{2}{\epsilon}$$ The result follows immediately.

One can begin with $$\frac{d}{dt}P \; \int_{-\infty}^\infty \frac{\phi(x)-\phi(t)}{x-t}dx.$$