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I've been experimenting with some differentiation, specifically the $x^x$ function. Using a table of standard derivatives, $\frac{d}{dx}x^n $ should be equal to $nx^{n-1}$. Using this logic, if $n = x$, then $\frac{d}{dx}x^x $ should be equal to $x \cdot x^{x-1}$ - which is equal to $x^x$. However, when I check with an online differentiation calculator, then it gives me the answer $x^x \cdot (\ln(x)+1)$. Why is it this and not $x^x$?

Gonçalo
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Mark Smith
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  • In order to find the desired result, you must make use of logarithmic differentiation. – Amy Skinner Oct 30 '23 at 23:07
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    "$\tfrac{\mathrm d}{\mathrm dx}(x^n) = nx^{n - 1}$" is a theorem which is true when $n$ is a constant, independent of $x$. It's the same as how for a constant $a$, we have $\tfrac{\mathrm d}{\mathrm dx}(ax) = a$, but it doesn't follow that $\tfrac{\mathrm d}{\mathrm dx}(x \cdot x) = x$ by letting $a = x$. – Izaak van Dongen Oct 30 '23 at 23:08
  • The rule $(x^n)' = n x^{n-1}$ only tracks the change in the base (and the source of the $+1$ in the answer you found), but the exponent is variable too. This is solely due to chain rule – Ninad Munshi Oct 30 '23 at 23:13

1 Answers1

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The rule $\frac{d}{dx}x^n=nx^{n-1}$ is under the assumption that $n$ is constant. (see power rule).

The answer you found online follows from exponent rules $$ a=c^{\log_c(a)}\iff a^b=(c^{\log_c(a)})^b=c^{b\log_c(a)}. $$ Let $a=x,b=x,c=e$ and we have $$ x^x=e^{x\ln(x)}. $$

Then, the derivative can be evaluated as $$ \begin{align} \frac{d}{dx}e^{x\ln(x)}&=e^{x\ln(x)}\frac{d}{dx}(x\ln(x))\\ &=e^{x\ln(x)}\left[\ln(x)\frac{d}{dx}x+x\frac{d}{dx}\ln(x)\right]\\ &=e^{x\ln(x)}(\ln(x)+1)\\ \frac{d}{dx}e^{x\ln(x)}&=x^x(\ln(x)+1). \end{align} $$

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