6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ A better notation would be $\frac{\partial }{\partial y}$, that is, you're looking at $f(x,y)=e^x$ so to speak. $\endgroup$ – Pedro Tamaroff Oct 3 '13 at 1:50
6
$\begingroup$

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 $y$).

$\endgroup$
  • 1
    $\begingroup$ Ok Thankyou thats a lot clearer now! :D $\endgroup$ – YYC Oct 3 '13 at 1:52
  • $\begingroup$ Just change $\frac d{dy}$ by $\frac d{dt}$... $\endgroup$ – Matemáticos Chibchas Oct 3 '13 at 2:36
  • $\begingroup$ Not sure what you mean $\endgroup$ – parsiad Oct 3 '13 at 3:42
  • $\begingroup$ "$x(y)$ is constant" and "$x$ does not depend on $y$" are not the same. The first implies that $x$ is the same regardless of the value of $y$, while the second implies $x$ is not a function of $y$ at all. I would suggest changing the wording there. $\endgroup$ – 6005 Nov 10 '14 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.