Problem with derivative of $x^{x^x}$

Question

I was recently watching blackpenredpen’s video (found here: https://m.youtube.com/watch?v=UJ3Ahpcvmf8) where he found the derivative of the the function $y = x^{x^x}$. Before watching the video, I decided to try it myself, and I didn’t it this way: Firstly I decided I would use the chain rule and define two functions, $u(x)$ and $v(x)$, both of which are $x^x$, so I could show the function as $u(v(x))$. Using the chain rule, the derivative would be: $u’(v(x))\cdot v’(x)$. We can find the derivative of $x^x$ (as shown on Quora here: https://www.quora.com/What-is-the-derivative-of-x-x-How-could-I-derive-it-as-well) as $x^x\cdot(\ln x+1)$. Now using the chain rule to get $u’(v(x))$ we get ${(x^x)}^{(x^x)}\cdot(\ln {x^x}+1)$ or $x^{x^{x+1}}\cdot(\ln {x^x}+1)$. This wasn’t the answer that blackpenredpen arrived at though. In the video, he arrived at $x^x\cdot x^{x^x}\left({\frac{1}{x}+\ln x+(\ln x)^2}\right)$. I can’t see how he did it, and I’m sure that it is right. If anyone could confirm my answer or show where I went wrong it would be greatly appreciated. Thanks.

Answer 1

You can apply the method given on quora to your problem: $$\left(x^{x^x}\right)'=\left(e^{x^x\ln x}\right)'=e^{x^x\ln x}\cdot \left(x^x\ln x\right)'=\\ =e^{x^x\ln x}\cdot \left(e^{x\ln x}\ln x\right)'=\\ =e^{x^x\ln x}\cdot \left[\left(e^{x\ln x}\right)'\ln x+e^{x\ln x}\cdot \frac1x\right]=\\ =x^{x^x}\cdot \left[e^{x\ln x}\left(x\ln x\right)'\ln x+x^x\cdot \frac1x\right]=\\ =x^{x^x}\cdot \left[x^x\left(\ln x+1\right)\ln x+x^x\cdot \frac1x\right]=\\ =x^{x^x}\cdot x^x\left[\ln^2 x+\ln x+\frac1x\right].$$ P.S. I could not watch the referenced video on Youtube, because it is not opening here (it might be blocked).

Answer 2

For differentiating expressions of the form $u(x)^{v(x)},$ I've always found it easiest to take logarithms and then differentiate implicitly. However, rather than work with the more general form $y = u^v,$ I'll work with $y = x^U$ (i.e. use $x$ for the base), which is sufficient to deal with $x^{x^x}.$

Taking the logarithm of both sides gives

$$ y = x^U \;\; \implies \ln y = \ln\left(x^U \right) \;\; \implies \;\; \ln y = U \cdot \ln x $$

Now differentiate both sides with respect to $x$:

$$ \frac{d}{dx}\left(\ln y\right) \;\; = \;\; \frac{d}{dx}\left(U \cdot \ln x \right) $$

$$ \frac{y'}{y} \;\; = \;\; \frac{d}{dx}(U) \cdot \ln x \; + \; U \cdot \frac{d}{dx} (\ln x) \;\; = \;\; U'\ln x \; + \; \frac{U}{x} $$

$$ y' \;\; = \;\; y \cdot \left[ U'\ln x \; + \; \frac{U}{x} \right] \;\; = \;\; x^U \cdot \left[ U'\ln x \; + \; \frac{U}{x} \right] $$

Using this result for $U(x) = x,$ we get

$$ \left(x^x\right)' \;\; = \;\; x^x \cdot \left[ 1 \cdot \ln x \; + \; \frac{x}{x} \right] \;\; = \;\; x^x(\ln x + 1) $$

Using that same result for $U(x) = x^x$ along with what we just found for $\left(x^x\right)',$ we get

$$ \left(x^{x^x}\right)' \;\; = \;\; x^{x^x} \cdot \left[ \left(x^x\right)' \cdot \ln x \; + \; \frac{x^x}{x} \right] $$

$$ \left(x^{x^x}\right)' \;\; = \;\; x^{x^x} \cdot \left[ x^x(\ln x + 1) \cdot \ln x \; + \; \frac{x^x}{x} \right] $$

$$ \left(x^{x^x}\right)' \;\; = \;\; x^{x^x} \cdot x^x \cdot \left[ (\ln x + 1)\ln x \; + \; \frac{1}{x} \right] $$

$$ \left(x^{x^x}\right)' \;\; = \;\; x^{x^x} \cdot x^x \cdot \left[ (\ln x)^2 \; + \; \ln x \; + \; \frac{1}{x} \right] $$