# What does this $0$ mean in Wolfram|Alpha's application of the chain rule?

If I enter this command to differentiate $(2+3x)^4$ I get the step-by-step output displayed here. The relevant part of the output is as follows:

Possible derivation:
$\frac{d}{d x}\left((2+3x)^4\right)$

Using the chain rule, $\frac{d}{d x}\left((3x+2)^4\right) = \frac{d u^4}{d u}0$, where $u=3x+2$ and $\frac{d}{d u}\left(u^4\right)=4u^3$:
$4(2+3x)^3\left(\frac{d}{d x}(2+3x)\right)$

I picked here a very simple example that could easily be done by hand. However, Wolfram|Alpha would use the same notation even if I picked a more complex example. I understand in basic terms what needs to be done to reach a solution, but I don't understand what the $0$ is doing in the notation.

• It's curious. Actually it shoud be $\frac{du}{dx}.$ – mfl Dec 3 '16 at 7:38

The actual calculation on Wolfram Alpha most probably is done by a Mathematica kernel in the background. My guess is that the $du/dx$ that is to be shown at that point is erroneously passed as expression to the kernel as something to evaluate. Trying to derive u rather than u[x], the kernel then interprets u as a constant, of which the derivative of course is $0$.