I know that the derivative of $\,e^x\,$ is $\,e^x$.
But how do I evaluate $\dfrac{d}{dx}{\large\left(e^{e^x}\right)}\,$?
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16$\begingroup$ Chain rule...${}$. $\endgroup$ – David Mitra Jul 17 '13 at 14:40
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2$\begingroup$ You need to use the chain rule. $\endgroup$ – James Jul 17 '13 at 14:41
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4$\begingroup$ What does this have to do with integration? $\endgroup$ – Chris Eagle Jul 17 '13 at 14:41
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$\begingroup$ Its related to Bell numbers I think so. $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jul 17 '13 at 14:57
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2$\begingroup$ It's funny that I asked the same question from my students last week. $\endgroup$ – Ali Jul 17 '13 at 16:52
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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:
$$\large \frac{d}{dx}\left(e^{(e^x)}\right) = \underbrace{e^x}_{f'(x)}\cdot\,\underbrace{e^{(e^x)}}_{e^{f(x)}}$$
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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
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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. $\endgroup$ – amWhy Jul 17 '13 at 15:35
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Hint: $$(e^u) '=u 'e^u$$
$$(e^{e^x}) '=e^xe^{e^x}$$
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take $u=e^x$ and $y = e^u$
$$ \large {y' = u'e^u = e^x e^{e^x}}$$
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Hint: Apply the chain rule. You would get $\frac d{dx}e^{e^x}=e^{x+e^x}$
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It's the derivative of a function of function. $\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$. So: $\frac{d}{dx}\exp(\exp(x))=\exp(x)\exp(\exp(x))$
answered Jul 17 '13 at 14:45
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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 $e^x$.
