Why does this differentation output correct result?

Question

I noticed something funny. If you differentiate $x^x$ treating the exponent as a constant, you get $xx^{x-1}=x^x$. If you treat the base as a constant, you get $x^x \ln{x}$. If you add these two bizzare and incorrect derivatives of $x^x$, you get $x^x(1+\ln{x})$, which is correct!

Is it merely a weird and funny coincidence or is it a part of a deeper result?

(I'm checking it for $(x^x)^x$, but it will take a while :) )

This website says: Note that the function defined by $y = x^x$ is neither a power function of the form $x^k$ nor an exponential function of the form $b^x$ etc... — Weather Vane, Aug 21 '17 at 18:52
I know that! That's why I find this result funny, you apply incorrect method twice to get a correct result — user473686, Aug 21 '17 at 18:52

score 6 · Answer 1 · edited Aug 21 '17 at 19:14

6

This is a general phenomenon. In this case $$x^x=f(x,x)$$ where $$f(u,v)=u^v.$$ By the chain rule, $$\frac{d}{dx}(x^x)=\frac d{dx}f(x,x)=f_1(x,x)+f_2(x,x)$$ where $f_1$ and $f_2$ are the partial derivatives of $f$. In this case $$f_1(u,v)=v u^{v-1}$$ (considering $v$ as constant) and $$f_2(u,v)=(\ln u) u^v$$ (considering $u$ as constant). So $f_1(x,x)=x^x$ and $f_2(x,x)=(\ln x)x^x$.

edited Aug 21 '17 at 19:14

Chappers

67,606

answered Aug 21 '17 at 19:00

Angina Seng

158,341

That looks like it generalises, allowing calculation of $\frac{d}{dx}(x\uparrow\uparrow n)$, ($n\in\Bbb N$), right? – Shuri2060 Aug 21 '17 at 19:08

Why does this differentation output correct result?

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