12
$\begingroup$

I know that the derivative of $\,e^x\,$ is $\,e^x$.

But how do I evaluate $\dfrac{d}{dx}{\large\left(e^{e^x}\right)}\,$?

$\endgroup$

closed as off-topic by José Carlos Santos, Kemono Chen, Gibbs, YiFan, mrtaurho Feb 11 at 15:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Kemono Chen, Gibbs, YiFan, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 16
    $\begingroup$ Chain rule...${}$. $\endgroup$ – David Mitra Jul 17 '13 at 14:40
  • 2
    $\begingroup$ You need to use the chain rule. $\endgroup$ – James Jul 17 '13 at 14:41
  • 4
    $\begingroup$ What does this have to do with integration? $\endgroup$ – Chris Eagle Jul 17 '13 at 14:41
  • $\begingroup$ Its related to Bell numbers I think so. $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jul 17 '13 at 14:57
  • 2
    $\begingroup$ It's funny that I asked the same question from my students last week. $\endgroup$ – Ali Jul 17 '13 at 16:52
27
$\begingroup$

To differentiate $\large e^{e^x},\,$ we use the chain rule.

$$\large \frac{d}{dx}\left(e^{f(x)}\right) = f'(x)\cdot e^{f(x)}$$

Here, we have that $e^{f(x)} = e^{e^x}$, so $f(x) = e^x$.

Thus $f'(x) = e^x,\,$ as you know. That gives us:

$$\large \frac{d}{dx}\left(e^{(e^x)}\right) = \underbrace{e^x}_{f'(x)}\cdot\,\underbrace{e^{(e^x)}}_{e^{f(x)}}$$

$\endgroup$
  • 1
    $\begingroup$ ${+1}{}{}{}$ TU! $\endgroup$ – mrs Jul 17 '13 at 14:53
  • 2
    $\begingroup$ All that large mathematics is ugly, but it is sadly true that without it, the exponents tend to get too small on screen. Is it generally better in such a case to switch notation, to $\exp e^x$, or is it confusing to mix those? $\endgroup$ – dfeuer Jul 17 '13 at 15:33
  • 3
    $\begingroup$ I've debated that @dfeuer. In general $\exp e^x$ would be the route I would go, in my own work. But in a situation like this, where the OP seems to clearly be confused, I'm afraid that to do so would be confusing to him/her. Thanks for the edit, btw. $\endgroup$ – Namaste Jul 17 '13 at 15:35
19
$\begingroup$

Hint: $$(e^u) '=u 'e^u$$

$$(e^{e^x}) '=e^xe^{e^x}$$

$\endgroup$
  • $\begingroup$ how do you add the spoiler thing? $\endgroup$ – AvatarOfChronos Jul 17 '13 at 15:11
  • 2
    $\begingroup$ @AvatarOfChronos You use >! $\endgroup$ – Pedro Tamaroff Jul 17 '13 at 15:28
  • 4
    $\begingroup$ $\large{\color{red}{+1}}$ for the spoiler trick! $\endgroup$ – Ali Jul 17 '13 at 16:51
  • $\begingroup$ @Ali: thx dear ali $\endgroup$ – M.H Jul 17 '13 at 18:38
  • 4
    $\begingroup$ Hit the "edit" button to see how it was done. (Then be sure to "cancel" rather than actually editing.) $\endgroup$ – GEdgar Jul 18 '13 at 1:32
14
$\begingroup$

take $u=e^x$ and $y = e^u$

$$ \large {y' = u'e^u = e^x e^{e^x}}$$

$\endgroup$
8
$\begingroup$

Hint: Apply the chain rule. You would get $\frac d{dx}e^{e^x}=e^{x+e^x}$

$\endgroup$
7
$\begingroup$

It's the derivative of a function of function. $\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$. So: $\frac{d}{dx}\exp(\exp(x))=\exp(x)\exp(\exp(x))$

$\endgroup$
6
$\begingroup$

Here's another method: for any positive function $f$, its derivative equals the function $f$ times its logarithmic derivative. In our case $f(x)=e^{e^x}$, so its logarithm, $e^x$, has derivative $e^x$.

$\endgroup$

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