So I'm going through the examples in my Diff. Eqs. book before I tackle the homework (Boyce/DiPrima) and in 1.2 Example 1 they invoke the chain rule to go from $\frac{dp/dt}{p-900}=\frac{1}{2}$ to $\frac{d}{dt}\ln|p-900|=\frac{1}{2}$.
I consider myself pretty familiar with the chain rule but I can't follow this step.
One way you can look at it is by integrating and differentiating the LHS (in that order). You probably know that for a function $f(x)$, $$f(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left[\int f(x)\,\mathrm{d}x\right]$$ So given $$\begin{align*} \frac{\frac{\mathrm{d}p}{\mathrm{d}t}}{p-900}&=\frac{1}{2}\\[1ex] \frac{\mathrm{d}}{\mathrm{d}t}\left[\int\frac{\frac{\mathrm{d}p}{\mathrm{d}t}}{p-900}\,\mathrm{d}t\right]&=\frac{1}{2}\\[1ex] \frac{\mathrm{d}}{\mathrm{d}t}\left[\int\frac{\mathrm{d}p}{p-900}\right]&=\frac{1}{2}\\[1ex] \frac{\mathrm{d}}{\mathrm{d}t}\ln|p-900|&=\frac{1}{2} \end{align*}$$ where the second to last line's integral can be approached via a substitution, say $u=p-900$, which is essentially the reverse of the chain rule for differentiation.