# Evaluating the derivative of $\large \;e^{e^x}$? [closed]

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

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

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

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

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:

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

• ${+1}{}{}{}$ TU! – mrs Jul 17 '13 at 14:53
• 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? – dfeuer Jul 17 '13 at 15:33
• 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. – Namaste Jul 17 '13 at 15:35

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

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

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

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

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

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

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

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 $e^x$.