TITLE: Math Joke about Differentiation
QUESTION [8 upvotes]: So I recently read a joke that goes like this:

A constant function and $e^x$ are walking on Broadway. Then suddenly the constant function sees a differential operator approaching and runs away. So $e^x$ follows him and asks why the hurry. "Well, you see, there's this differential operator coming this way, and when we meet, he'll differentiate me and nothing will be left of me...!" "Ah," says $e^x$, "he won't bother ME, I'm e to the x!" and he walks on. Of course he meets the differential operator after a short distance.
ex: "Hi, I'm $e^x$"
diff.op.: "Hi, I'm $d\over dy$"

What I dont understand is, Isn't ${d\over dy}\left(e^x\right)$ supposed to be
$e^x {dx\over dy}$ because of the Chain rule? So what does the joke mean?

REPLY [6 votes]: With respect to $y$, $f\left(x\right)\equiv e^x$ is constant (changing $y$ will not change $f$!). In other words,
$$\frac{d}{d y}\left(e^{x}\right)=0$$
Extra clarification:
I think what you are calling "implicit differentiation" is the
chain rule. If $x$ depends on $y$, (write this as $x\left(y\right)$)
$$
\frac{\partial}{\partial y}\left(e^{x\left(y\right)}\right)=e^{x\left(y\right)}x^{\prime}\left(y\right).
$$
Of course, the joke was assuming that $x\left(y\right)$ is constant
(i.e. it does not depend on $y$).