Differentation with respect to what?

Question

I saw in a book that differentiated the volume of a cylinder $V= \pi r^2 l$ like this.

\begin{align*} V &= \pi r^2 l\\ \implies \ln(V) &= \ln\left(\pi r^2 l\right)\\ &= \ln(\pi)+\ln\left(r^2\right)+\ln(l)\\ &= \ln(\pi)+2\ln(r)+\ln(l)\\ \implies \frac{dV}{V} &= 0+2\frac{dr}{r}+\frac{dl}{l} \end{align*}

Where $r$ is the radius of the cross-section of the cylinder's base, and $l$ is the height of the cylinder.

At first, they took the natural logarithm on both sides and then differentiated it (I don't know w.r.t. what), and then they got this result.

Can someone please explain how this is done?

Answer 1

It's a shortcut.

$V=\pi r^2l$

$dV=\frac{\partial V}{\partial r}dr + \frac{\partial V}{\partial l}dl$

$dV=2\pi r l dr + \pi r^2dl$

$\frac{dV}{V}=\frac{2}{r}dr+\frac{dl}{l}$

Answer 2

They're not differentiating with respect to anything; that is, they're not calculating the derivative with respect to anything. Instead, they're just differentiating; that is, they're calculating the differential. If you want to know about the derivative with respect to something (say time $ t $), then you divide by its differential (say, divide by $ \mathrm d t $), and now you've differentiated with respect to something. But until then, it's just the differential, which is absolute.

Formally, you can think of the differential as a derivative with respect to an unspecified variable; or treat $ V $, $ l $, and $ r $ as names for functions and interpret $ \mathrm d V $ as $ V ' $ etc. But in applied math, we usually don't worry about this formality. (As long as somebody has made it rigorous at some point, you know that you can just do it.)

Note that if you combine the Chain Rule with the fact that the derivative of the natural-logarithm function is the reciprocal function, you find that the derivative of $ \ln u $ (for any differentiable quantity $ u $) with respect to $ x $ (for any independent variable $ x $) is $ \frac { \mathrm d } { \mathrm d x } ( \ln u ) = \frac 1 u \, \frac { \mathrm d u } { \mathrm d x } $. Now to make this not refer to $ x $ at all, you multiply both sides by $ \mathrm d x $ to get the Natural-Logarithm Rule for differentials: $ \mathrm d ( \ln u ) = \frac 1 u \, \mathrm d u $. This is the rule that's being applied, in three separate places, in the last step. If you think of every special function as giving you a rule for differentials, rather than a derivative, then you almost never have to apply the Chain Rule explicitly (because it's been incorporated into all of the various rules for differentials.)