Why can't we differentiate $x^x$ like we differentiate $x^3$ (e.g.)?

Question

I know the derivative of $x^x$ is $x^x(1+log(x))$. So that must mean we cannot differentiate it like we do, say $x^5$.

The reason I read explaining why this is so is that the proof of the formula that the derivative of $x^n$(where n is a constant) $=$ $n.x^{n-1}$ involves binomial theorem.

But when I read the proof, I couldn't identify why there'd be any difference in applying that theorem for $x^x$.

So, I want to ask why can't we use the binomial theorem for justifying the same formula for, $x^x$ or $a^x$?

EDIT: I am explicitly asking why the 'binomial theorem proof' doesn't act as a proof for $x^x$, $a^x$ and so on. I appreciate the points given in the already existing answers, but I want a direct answer to my actual question? Thank you.

Answer 1

Learning formulas by heart is OK... when they work. When they do not, and we wonder why, we must go back to the formal definition: $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$ So, when $f(x)=x^n$, you write: $$f'(a)=\lim_{h\to 0}\frac{(a+h)^n-a^n}{h}=\cdots$$ and some tricks allow me to simplify this. But note that in the subtraction, both terms have the same exponent $n$.

Now, if $f(x)=x^x$, write: $$f'(a)=\lim_{h\to 0}\frac{(a+h)^{a+h}-a^a}{h}$$ and there is no direct trick to compute this easily. So you must resort to the logarithm (which is indeed related to the definition of $x^x$) and the chain rule.

Answer 2

Differentiation is a simple procedure and it starts with a very simple definition and leads to laws of differentiation : sum/differencr, product, quotient, chain rule, implicit differentiation. Another part is establishing the derivatives of $x^{n}$, circular functions (and their inverses), exponential and logarithmic functions. These are building blocks of what we call elementary functions which are built using these blocks via a finite number of algebraic operations and a finite number of compositions.

Thus it is clear that using the derivatives of these building blocks and the rules of differentiation one can evaluate the derivative of any elementary function in an almost mechanical fashion. The function under consideration $f(x) =x^{x} $ involves base and exponent and we see that there are two building blocks which look similar to it namely $$(x^{n}) '=nx^{n-1},\,(a^{x})'=a^{x}\log a$$ Note that in both these formulas either the base or the exponent is a constant and hence these formulas can't be used directly to deal with $f$ where both the base and exponent are variable. Why don't we include functions like $f$ (where both base and exponent are variable) in the list of building blocks for elementary functions? Because it is unnecessary. This function $f$ can be expressed using the existing building blocks and composition. Thus $x^{x} =e^{x\log x} $ and we can apply the rules now to get its derivative.

Also note that the building blocks are defined via very simple formulas so that their derivatives can be easily evaluated using the limit definition. It is simply not possible to apply the limit definition to evaluate derivative of $f(x) =x^{x} $ in a manner which does not mirror the application of chain rule.