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!

Got it, thank you! Is there any way to correct method 2 or solve the problem without logarithmic differentiation? — EthanAlvaree, May 07 '21 at 06:19
I don't think so. The most natural way to define the expression $f(x)^{g(x)}$ is to write $f(x)^{g(x)}:=\exp\left[g(x)\ln f(x)\right]$, so logarithmic differentiation seems inevitable. — Alann Rosas, May 07 '21 at 06:22
Please do not modify your question after you have received an answer. — Parcly Taxel, May 07 '21 at 06:32
I think you can proceed as in @AlannRosas comment by writing your function as $\exp(a^x\log x) $. — Paramanand Singh, May 07 '21 at 06:32
Also I don't see how the chain rule can be applied here unless we use $\exp, \log$. — Paramanand Singh, May 07 '21 at 06:35

score 3 · Answer 1 · answered May 07 '21 at 06:05

3

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.

answered May 07 '21 at 06:05

Parcly Taxel

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Got it, thank you! Is there any way to correct method 2, or solve the problem without logarithmic differentiation? – EthanAlvaree May 07 '21 at 06:18
@Mathemanic Please ask that in another question. – Parcly Taxel May 07 '21 at 06:21
Edited original post to clarify question and intent. – EthanAlvaree May 07 '21 at 06:29

score 2 · Accepted Answer · edited May 07 '21 at 17:29

2

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.

edited May 07 '21 at 17:29

EthanAlvaree

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answered May 07 '21 at 06:37

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Thank you! Do you know the name of this "generalized" version of chain rule? Or could you point me in the direction of where I could find a proof of this generalized form of chain rule? – EthanAlvaree May 07 '21 at 06:38
Look in the index of your calculus book for this: "Chain rule for partial derivatives" – GEdgar May 07 '21 at 17:55

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

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