Derivative of ${ x }^{ x }$ without logarithmic differentiation

Question

With logarithmic differentiation, it is quite simple to compute the derivative of $x^x$:

$$y=x^x$$ $$\ln {y} =x \ln{x}$$ $$\frac {1}{y} \frac {dy}{dx} = \ln{x} +1$$ $$\frac {dy}{dx} =\left( ln(x)+1 \right) x^x.$$

Is there a method to compute the derivative of ${ x }^{ x }$ that does not rely on logarithmic differentiation?

Sort of: you can rewrite $x^x$ as $e^{x\ln x}$ and differentiate that. But the computations are essentially identical. — Brian M. Scott, Mar 24 '13 at 03:01
Thanks for the comment, I should have thought of that! I'll type up an answer. — Gamma Function, Mar 24 '13 at 03:03
You should be able to do it via first-principles right? Although its definitely not a better method — user71207, Jan 05 '21 at 06:34

score 10 · Accepted Answer · edited Mar 24 '13 at 03:51

10

Thanks to Brian M. Scott for the comment that led to this solution:

$y={ x }^{ x }={ e }^{ x\ln { x } }\\ \frac { dy }{ dx } =\left( \ln(x)+1 \right) { e }^{ x\ln { x } }\\ \frac { dy }{ dx } =\left( \ln(x)+1 \right) { x }^{ x }\\ $

edited Mar 24 '13 at 03:51

Brian M. Scott

616,228

answered Mar 24 '13 at 03:07

Gamma Function

5,642

7

Logarithmic differentiation and this one is essentially the same. – hrkrshnn Jun 13 '13 at 14:42

score 9 · Answer 2 · answered Mar 24 '13 at 04:41

9

There's another way that looks like a gross blunder but actually is perfectly correct. I will illustrate it on the more general question, differentiate $y=f(x)^{g(x)}$.

If $g$ were constant, we'd get $f'(x)g(x)f(x)^{g(x)-1}$.

If $f$ were constant, we'd get $g'(x)f(x)^{g(x)}\log f(x)$.

Add these together to get the answer: $$y'=f'(x)g(x)f(x)^{g(x)-1}+g'(x)f(x)^{g(x)}\log f(x)$$

It's easy to see that in the original problem, where $f(x)=x$ and $g(x)=x$, this reduces to $x^x+x^x\log x$, as obtained by other methods.

answered Mar 24 '13 at 04:41

Gerry Myerson

179,216

Why did you add them together? – A Googler Jan 18 '15 at 11:06
@AGo, mostly because it works. – Gerry Myerson Jan 18 '15 at 17:24
Yeah , i meant why does it work? – A Googler Jan 19 '15 at 11:24
@AGo, I think it's an application of a version of the chain rule. If $y$ is a function of $s$ and $t$, each of which is a function of $x$, then $dy/dx=(\partial y/\partial s)(ds/dx)+(\partial y/\partial t)(dt/dx)$. Here we have $y=s^t$ with $s=f(x)$, $t=g(x)$. – Gerry Myerson Jan 20 '15 at 00:53
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@GerryMyerson: it is an application of the chain rule (no doubt about it) – robjohn Mar 31 '15 at 11:54
(+1) this is not logarithmic differentiation (or at least it is so deeply hidden that it is quite hard to see). To me, this answers the question better than differentiating $e^{x\log(x)}$. – robjohn Mar 31 '15 at 11:58
What's funny about this approach is that it's the sum of the very things we tell first-semester calculus students not to do when differentiating a variable raised to a function exponent. But in fact this is the correct "general rule" for which the "Power Rule" (variable raised to constant) and the "exponential function rule" (constant raised to variable) are special cases... – Dec 29 '22 at 01:01

Derivative of ${ x }^{ x }$ without logarithmic differentiation

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