Check My Work: Chain Rule doesn't work when differentiating $x^{\ln(x)}$ ??

Question

I am asked to find the derivative of

$$ f: (0, \infty) \to \mathbb{R},~x \mapsto x^{\ln(x)} $$

I am aware that I can rewrite it as $f(x) = e^{\ln^2(x)}$, and easily differentiate but I wanted to use the chain rule on the original function. Here's my work:

The chain rule states $$ \frac{d}{dx}~g(h(x)) = g'(h(x))\cdot h'(x) $$ In this example, the 'outer' function is $g(x) := x^x$ with $g'(x) = x^x (\log(x) + 1)$ and the 'inner' function is $h(x) := \ln(x)$ with $h'(x) = x^{-1}$.

Appling the chain rule from above, I obtain

$$ f'(x) = \ln(x) ^{\ln(x)}~(\ln(\ln(x) + 1 )~\frac{1}{x} $$

When I let WolframAlpha plot this, I recognised my solution to be vastly different from the proposed solution $$ f'(x) = \frac{2}{x}~\ln(x)~x^{\ln(x)} $$ Where have I gone wrong?

Answer 1

I assume you already saw your mistake as pointed in the comments and in @gimusi's answer. A different approach (and one I find useful) is to take take $\log$ of your exponential function:

Let $y=\log(x^{\log(x)})=\log^2(x)$ and note that $e^y=x^{\log(x)}$. Then you have that

$$\frac{dy}{dx}=\frac{2\log(x)}{x}$$

Hence you have that $$\frac{d(x^{\log(x)})}{dx}=\frac{d(e^y)}{dx}=e^y\frac{dy}{dx}=x^{\log(x)}\frac{2\log(x)}{x}$$

which is the answer you want

Answer 2

Your method is wrong since with

• $g(x)=x^x$
• $h(x)=\log x$

we have that

$$g(h(x))=(\log x)^{\log x}\neq x^{\log x}$$