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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Sign up to join this communityI know that the derivative of $\,e^x\,$ is $\,e^x$.
But how do I evaluate $\dfrac{d}{dx}{\large\left(e^{e^x}\right)}\,$?
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:
$$\large \frac{d}{dx}\left(e^{(e^x)}\right) = \underbrace{e^x}_{f'(x)}\cdot\,\underbrace{e^{(e^x)}}_{e^{f(x)}}$$
Hint: $$(e^u) '=u 'e^u$$
$$(e^{e^x}) '=e^xe^{e^x}$$
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 $e^x$.