Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

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

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

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – anorton, user72694, Davide Giraudo, Antonio Vargas, Mark Bennet
If this question can be reworded to fit the rules in the help center, please edit the question.

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.