Understanding the step from $x_A\frac{d}{dx_A}$ to $\frac{d}{d \ln(x_A)}$

Question

I've come across this step as part of some chemistry I'm doing, and I don't simply know why this step was able to be done. $x_A$ is mole fraction of a species "A", and $x_A = \frac{C_A}{C_o}$

The whole thing done out is:

$\frac{\partial \mu_A}{\partial x_A} = RT\frac{\partial}{\partial x_A}(\ln(\gamma_A)+\ln(x_A))$

$\frac{\partial \mu_A}{\partial x_A} = \frac{RT}{x_A}(x_A\frac{\partial \ln(\gamma_A)}{\partial x_A}+1)$

$\frac{\partial \mu_A}{\partial x_A} = \frac{RT}{x_A}(\frac{\partial \ln(\gamma_A)}{\partial \ln(x_A)}+1)$

Answer 1

I am going to write some symbols out the way I have seen similar notation manipulated, but I have never liked what I am about to write and take no responsibility for it: $$ x D\ln (f(x)) = x\dfrac{f'(x)}{f(x)} = \dfrac{df(x)}{dx} \dfrac{x}{f(x)}= \dfrac{df(x)/f(x)}{dx/x}= \dfrac{d\ln(f(x))}{d\ln(x)}. $$

Answer 2

Let $y=\ln x$. Using the chain rule, $\dfrac d{dx}=\dfrac d{dy}\dfrac {dy}{dx}=\dfrac1x\dfrac d{dy}$, so $x\dfrac d{dx}=\dfrac d{dy}$.