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I am asked to find the derivative of $\left(x^x\right)^x$. So I said let $$y=(x^x)^x \Rightarrow \ln y=x\ln x^x \Rightarrow \ln y = x^2 \ln x.$$Differentiating both sides, $$\frac{dy}{dx}=y(2x\ln x+x)=x^{x^2+1}(2\ln x+1).$$

Now I checked this answer with Wolfram Alpha and I get that this is only correct when $x\in\mathbb{R},~x>0$. I see that if $x<0$ then $(x^x)^x\neq x^{x^2}$ but if $x$ is negative $\ln x $ is meaningless anyway (in real analysis). Would my answer above be acceptable in a first year calculus course?

So, how do I get the correct general answer, that $$\frac{dy}{dx}=(x^x)^x (x+x \ln(x)+\ln(x^x)).$$

Thanks in advance.

Zugzwang14
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  • If $x<0$ then $x^x$ is not well-defined in general. – Thomas Andrews Jun 01 '13 at 16:47
  • So why does Wolfram Alpha have the answer I give at the end of my post? http://www.wolframalpha.com/input/?i=derivative+of+%28x%5Ex%29%5Ex – Zugzwang14 Jun 01 '13 at 16:49
  • Not clear what you mean. WA is telling you that $x<0$ is not allowed, and also that $x=0$ is not allowed. (Why $x=0$ is not allowed is a little different.) It's a little weird that WA graphs the function for $x\leq 0$, however. There, it is graphing complex values, but it makes the output a mess. – Thomas Andrews Jun 01 '13 at 16:50
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    Actually, my question was completely stupid. Even though I found the derivative, I then failed to recognise that my answer is identical to the one WA gives because they are written in a different form. In any case, thanks! – Zugzwang14 Jun 01 '13 at 16:54
  • @ThomasAndrews See this question of mine. – Lee Sleek Jun 01 '13 at 21:58

2 Answers2

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in step $$\dfrac{dy}{dx}=y(2x\ln x+x)$$ $$\dfrac{dy}{dx}=y(x\ln x+x\ln x+x)$$ in question $y=(x^x)^x$ $$\dfrac{dy}{dx}=(x^x)^x(x\ln x+\ln x^x+x)$$ just rearrange your second last step

iostream007
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If $y=(x^x)^x$ then $\ln y = x\ln(x^x) = x^2\ln x$. Then apply the product rule:

$$ \frac{1}{y} \frac{dy}{dx} = 2x\ln x + \frac{x^2}{x} = 2x\ln x + x$$

Hence $y' = y(2x\ln x + x) = (x^x)^x(2x\ln x + x).$

This looks a little different to your expression, but note that $\ln(x^x) \equiv x\ln x$.

Fly by Night
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    And $x^{x^2+1} = (x^x)^x \cdot x$. – Thomas Andrews Jun 01 '13 at 16:53
  • @ThomasAndrews I don't follow, sorry. – Fly by Night Jun 01 '13 at 16:59
  • @FlybyNight To multiply powers with the same base, add the exponents. The power of a power law means left -associative exponentiation is actually multiplication of the exponents. $x$ can be written as $x^1$, and $(x^x)^x$ can be written as $x^{x^2}$. Multiplying them results in adding the exponents and producing the desired answer. – Lee Sleek Jun 01 '13 at 22:01
  • @LeeSleek Thanks for reminding me how manipulate indices. I was asking Thomas for clarification of the reason behind his post. The tone implied that I had made an error, which I had not. – Fly by Night Jun 01 '13 at 22:46
  • @FlybyNight I believe he was suggesting that you not leave the answer in factored form (i.e. write it as $2x^{x^2 + 1} \ln x + x^{x^2 + 1}$). – Lee Sleek Jun 02 '13 at 01:22