On the mass conservation equation of stellar structure

In class, this perhaps was not completely transparent.  Let's try again.  Consider a spherical shell of thickness dr located at radius r from the center of a sphere.  It will have surface area $4\pi r^2$ and we can, since dr is very small, get its volume just by taking it to be a rectangular prism with area $4\pi r^2$ and thickness dr, so its volume is
$$dV=4\pi r^2 dr.$$

At a radius r, the star has density $\rho (r )$, so the differential mass enclosed by this shell is just
$$dM=\rho( r)dV=4\pi r^2 \rho (r ) dr.$$

Dividing both sides through by dr, we have the mass conservation equation:
$$\frac{dM}{dr}=4\pi r^2 \rho (r ).$$

Note that if we integrate from zero up to radius R, we get the total mass within radius R:

$$M (<R )=\int_0^R 4\pi r^2 \rho ( r) dr.$$

Note also that differentiating this integral with respect to R will give
$$\frac{dM}{dR}=4\pi R^2\rho ( R)$$
by the fundamental theorem of calculus!

In class, Professor Johnson argued that we have
$$M=\frac{4}{3}\pi r^3\rho ,$$ so we can differentiate to get
$$\frac{dM}{dr}=4\pi r^2 \rho .$$

This is not quite right.  Mass only equals $\frac{4}{3}\pi r^3 \rho$ if $\rho$ is the average density of the star, and that is not a function of radius: it is just total mass divided by total volume.  So we do not recover that we must evaluate the density at a specific radius in the mass conservation equation if we approach it this way.  Also, if we did make $\rho$ a function of radius, then differentiation would lead to two terms by product rule, as Jimmy pointed out in class.  So this derivation is not the best way to think about mass conservation.