Differentiate $x^{a^x}$ without logarithmic differentiation?

Question

Problem. Compute $\frac{d}{dx} x^{a^x}$.

Method 1: Logarithmic Differentiation (Correct):

\begin{align*} y&= x^{a^x} \\ \ln(y)&=a^x \cdot \ln(x) \\ \frac{1}{y} \frac{dy}{dx} &=a^x \frac{1}{x} + \ln(x) \ln(a) a^x \\ \frac{dy}{dx} &= y\left(\frac{1}{x} a^x + \ln(a) \ln(x) a^x\right) \\ \frac{dy}{dx} &= x^{a^x}\left(\frac{1}{x} a^x + \ln(a) \ln(x) a^x\right) \\ \frac{dy}{dx} &= a^x x^{a^x}\left(\frac{1}{x} + \ln(a) \ln(x)\right) \\ \frac{dy}{dx} &= a^x x^{a^x-1}\left(1 + x\ln(a) \ln(x) \right) \\ \end{align*}

Method 2: Chain Rule (Incorrect):

\begin{align*} y&= x^{a^x} \\ \frac{dy}{dx} &= a^x x^{a^x-1} \cdot \left[\ln(a) \cdot a^x\right] \end{align*}

Method 2 is incorrect because it $x^{a^x}$ is not a power function, so we cannot apply power rule (thanks @Alann_Rosas and @Parcly_Taxel).

My Question: Can Ninad Munshi's answer here be adopted to correct method 2 without the use of logarithmic differentiation? What is the name (and/or proof) of this generalized version of chain rule?

Thank you!

Answer 1

As suggested, one can use the answer in Why the chain rule does not work for this question? to solve this problem.

\begin{align*} \frac{d}{dx} x^{a^x} &= a^x \cdot x^{a^x-1} + x^{a^x} \cdot \ln(x) \cdot \frac{d}{dx} a^x \\ &= a^x \cdot x^{a^x-1} + x^{a^x} \cdot \ln(x) \cdot \ln(a) \cdot a^x \end{align*}

This is a special case of the multivariable chain rule.

Answer 2

It is not a misapplication of the chain rule, but of the power rule. The power rule does not work for non-constant exponents; by that logic the derivative of $x^x$ would be $x\cdot x^{x-1}=x^x$, which is flat-out wrong.