# Problem involving implicit differentiation with many exponents. [closed]

Find the derivative of the function $x^{x^{x^{x}}}$.

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## closed as off-topic by anorton, user72694, Davide Giraudo, Antonio Vargas, Mark BennetAug 5 '14 at 16:29

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

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Let $f_n(x) = x\uparrow \uparrow n$. Note that $$f_n(x) = x^{f_{n-1}(x)}$$ $$\log(f_n(x)) = f_{n-1}(x) \log(x)$$ $$\dfrac{f_n'(x)}{f_n(x)} = f_{n-1}'(x) \log(x) + \dfrac{f_{n-1}}{x}$$ Note that $f_1(x) = x \implies f_1'(x) = 1$. Hence, $$f_2'(x) = f_2(x) \left(\log(x) + 1\right) = x^x (1 + \log(x))$$ Similarly, find $f_3'(x)$ and $f_4'(x)$ using the recurrence.

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Welcome to SE Mathematics! Just a courtesy note: please don't just say "Find..." or "Prove..." (for whatever reason, I think it's frowned upon) Also, we like to see what you have tried already, instead of just a problem statement.

That said: $$y = x^{x^{x^{x}}}$$ $$\ln(y) = x\ln(x^{x^{x}})$$ $$\ln(\ln(y)) = x\ln(x\ln(x^{x}))$$ $$\ln(\ln(\ln(y)))=x\ln(x\ln(x\ln(x)))$$

Now you have a chain rule on the left and a product/chain rule on the right.

If that seems hard, try to differentiate $x^x$ and $x^{x^x}$ first.

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