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Why can't I take the derivative of $x^x$ as $x(x^{x-1})$?

I don't understand why I have to convert it to $e^{x\ln(x)}$ first.

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    How do you propose to differentiate $x^{x-1}$? – 2017-01-04
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    The exponent in this case is a variable. The "power rule" works for fixed exponents. – 2017-01-04
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    @Dave Thanks, your comment clarified what I was looking at - I wasn't sure where the chain rule was supposed to be coming in. – 2017-01-04
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    Power rule doesn't work for changing exponents – 2017-01-04
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    Because the formula you want to apply is for the derivative of $x^{\text{a fixed exponent}}$. Here the exponent varies. – 2017-01-04
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    The main problem is... how will you use the chain rule? Chain rule works like this: derivative of a function of a single argument, times the derivative of the argument. Here, $x$ occurs in two places - where is the chain rule supposed to apply? You have to convert it into a form, where all $x$-related stuff is together, otherwise chain rule doesn't even come into play. That's why you put everything into the exponent. – 2017-01-04
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    @user402571 Have you tried evaluating the limit definition of the derivative? That might demonstrate why you're proposal doesn't work for $x^x $. – 2017-01-04
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    I mentioned the chain rule because you use it after you convert it. – 2017-01-04

7 Answers 7

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You want to differentiate $x\mapsto x^x$ with respect to "$x$".

You can not use the rule

$$\frac{\mathrm d}{\mathrm d x}(x^n)=nx^{n-1}$$

because here, "$n$" is depending on $x$.

What you are stated is correct, you have to use the identity

$$x^x=e^{x\ln(x)}$$

and then use the chain rule.

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    "The process of finding a derivative is called differentiation." – 2017-01-04
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Taking the basis as variable, the derivative is $x\cdot x^{x-1}$, taking the exponent as such the derivative is $x^x\cdot\ln(x)$. Adding both gives the derivative of $x^x$, namely $$\left(x^x\right)'=x^x+x^x\cdot\ln(x).$$ Astonishing at the first glance, isn't it?

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I think you are asking why you can't differentiate $x^x$ in the same way you would differentiate $x^3$?

The shortcut that you use when differentiating $x^r$ where $r$ is an real number, only hold when $r$ is a real number. In the case of $x^x$, the exponent is a variable so the shortcut doesn't apply.

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    But when you write $x^x$ as $e^{xln(x)}$, x is still in the exponent – 2017-01-04
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    @user402571 Yes, but instead of $f(x)^{f(x)}$, it's now in the form of $c^{f(x)g(x)}$. – 2017-01-05
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    How would I know that something is in a form is directly differentiable? – 2017-01-05
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just hint

It doesn't work cause the exponent is not constant.

we cannot factor $x^x-x_0^{x_0}$

it is not like $x^n-x_0^n$

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    But when you write $x^x$ as $e^{xln(x)}$, x is still in the exponent. – 2017-01-04
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Some people have given good hints, but I would like to elaborate on their responses.

Students get confused with the power rule very often. I've seen these at all levels. The key to remember is that the power rule allows us to look at derivatives of polynomials - specifically, those functions whose powers are fixed numbers.

We know, and can show through induction (or even the binomial theorem; do it!) that

$$ \frac{d}{dx} [x^n] = nx^{n-1} $$

where $n \in \mathbb{R}$. What you have is a function whose base and power both vary. To use a rule we know, we must do the typical $e$ trick: recall that

$$ x^n = e^{n \ln x} $$

and so we just write $ x^x = e^{x \ln x}$ and use our typical differentiation rules.

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This is a valid assumption many students make, but the reason is because $x^x$ is not like the power rule we have seen. The power rule applies to functions like $5x^4, 3x^2, x^{3.2}... etc$. Notice the pattern? The $exponent$ is a $number$ that stays $constant$. However, something like $x^x$, is not. The $x$ in the exponent is always $changing$, that's what $x$ does. So you cannot.

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As people have said, you cannot use the power rule to differentiate non-constant powers; it only works for functions resembling $f(x)^c$ where $c$ is a constant.

To differentiate $y = x^x$...

$ln(y) = ln (x^x)$

$ln(y) = x ln(x)$

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

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

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

$\frac{dy}{dx} = (ln(x)+1)x^x$

$\frac{dy}{dx} = x^xln(x)+x^x$