# Computing derivative $x^{x^x}$

Could you show me how to compute $f'(x)$, where $f(x)=x^{x^x}$.

I know that for $g(x)=x^x=e^{x\ln x} \ \$ $g'(x)=e^{x\ln x}(\ln x+1)$

Now, my problem is this: is $f(x)=x^{x^x}= e^{x^x \ln x}$ or $e^{x\ln x^x}$ ?

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Assuming you meant $x^{x^x}=x^{(x^x)}$, then $f(x)=e^{x^x\ln x}$ – Rick Decker Jan 18 '13 at 16:13

## 3 Answers

With $g(x)=x^x$, you can write $f(x)=x^{g(x)}=e^{g(x)\ln{x}}$. We now have $f'(x)=e^{g(x)\ln{x}}(g'(x)\ln{x}+g(x)\cdot\frac{1}{x})$, and you can just insert the expressions for $g(x)$ and $g'(x)$ and simplify :)

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Since you know the derivative of $g(x)=x^x$, you can use the chain rule.

Then we can write $x^{x^x}=x^{g(x)}= e^{\ln (x^{g(x)})}=e^{g(x) \ln x}$.

Now use the chain rule (twice), and compute.

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Writing $x^{x^x}$ is meant as $x^{(x^x)}$, therefore $$x^{x^x} = e^{x^x\ln x}$$ Once you know that $(x^x)'=x^x(1+\ln x)$, you have $$D[x^{x^x}] = x^{x^x}\cdot D[x^x\cdot\ln x] = x^{x^x+x-1}\big( x\ln x + x(\ln x)^2 + 1\big)$$

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