Derivative of $x^x$ and the chain rule

Question

Rewriting $x^x$ as $e^{x\ln{x}}$ we can then easily calculte the ${\frac{x}{dx}}$ derivative as ${x^x}(1 + \ln{x})$. We need to use chain rule in form $\frac{de^u}{du}\frac{du}{dx}$. The question is why cannot we use the chain rule skipping the 1st step of rewriting $x^x$ as $e^{xlnx}$? The idea would be to write $\frac{u^x}{du}\frac{du}{dx}$ which simply would be $x^x\ln{x} \cdot 1$? The answer is incorrect, here we are missing the $+x^x$ term.

Answer 1

$$\frac{d(a^x)}{dx}=a^x\ln a$$ is true only for constant $a\gt0$.

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Note that in getting the above result, the steps followed are similar to the steps for getting the derivative of $x^x$.

Let $f(x)=a^x$, $a\gt0$. Then, $$f(x)=e^{x\ln a}$$ $$f'(x)=\frac{d(e^{x\ln a})}{d(x\ln a)}\cdot\frac{d(x\ln a)}{dx}=e^{x\ln a}\cdot\ln a=a^x\ln a$$

whereas, for $f(x)=x^x$, $x\gt0$, $$f(x)=e^{x\ln x}$$ $$f'(x)=\frac{d(e^{x\ln x})}{d(x\ln x)}\cdot\frac{d(x\ln x)}{dx}=e^{x\ln x}\cdot(\ln x+1)=x^x(\ln x+1)$$