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I was killing time in a meeting where I wasn't needed and tried to calculate $\tfrac{\mathrm{d}}{\mathrm{d}x}\left[e^{x^x}\right]$. I already know I'm wrong from looking up the answer, but I'd like to know where I went amiss, if you had a minute.

My attempt: $$\begin{align}y &= e^{x^x} \\ \ln{y} &= x\ln{e^x} \\ \ln{y} &= x^2 \\ \frac{dy}{dx}\frac{1}{y} &= 2x \\ \frac{dy}{dx} &= 2xy \\ &= 2xe^{x^x} \end{align}$$ I was able to use this cool differentiation site to see a step-by-step answer using the "generalized power rule" (which I was never taught or figured out) giving the correct answer of $$\frac{\mathrm{d}}{\mathrm{d}x}\left[e^{x^x}\right]=x^x\mathrm{e}^{x^x}\left(\ln\left(x\right)+1\right)$$ but to me the approach of taking natural logs of both sides should work. So I obviously went wrong somewhere.

Thanks!

Charlie
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3 Answers3

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If $y=e^{x^x}$, then $\log(y)=x^x\ne x\log(e^x)=\log(e^{x^2})$

Note that $e^{x^x}\ne (e^x)^x=e^{x^2}$.

So, to differentiate $y$, we use $\log(\log(y))=x\log(x)$. Then,

$$\frac{d\log(\log(y))}{dx}=\log(x)+1=\frac{1}{y\log(y)}\frac{dy}{dx}$$

whence solving for $\frac{dy}{dx}$ and using $y=e^{x^x}$ and $\log(y)=x^x$ yields the coveted result.

Mark Viola
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$f(x)=e^{g(x)}$ where $g(x)=x^x=e^{xln(x)}$, $f'(x)=e^{g(x)}g'(x)$, $g'(x)=e^{xln(x)}(ln(x)+1)$.

Your mistake is when you compute $ln(y)$.

Tsemo Aristide
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Note that $ln (y) = x^x$ and not equal to $x ln (e^x) $

Thsi is because $ln (e^(x^x))$ cancels the $e $ and the exponent $x^x $ is tue result, from there derivate as normal and obtain the result.

Jonathaniui
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