Prove $\frac{d}{d x} \ln |a x|=\frac{d}{d x} \ln |x|$ for any constant a.

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Given: $a$ is a constant

To proof: $\frac{d}{dx}\ln |ax|=\frac{d}{dx}\ln |x|$

$\textbf{PROOF}$

Since the natural logarithm $\ln x$ is only defined when the argument $x$ is positive, it is safe to assume that $a$ and $x$ are both positive without loss of generality (such that $|x|=x>0$ and $|ax|=ax>0$ ).

\begin{align*} \frac{d}{dx}\ln |ax|&=\frac{d}{dx}\ln (ax) \\ &\color{#4257b2}\text{Use the propert of logarithms: }\ln(ab)=\ln a+\ln b \\ &=\frac{d}{dx}\left(\ln a+\ln x\right) \\ &\color{#4257b2}\text{Derivative of sum is sum of derivatives} \\ &=\frac{d}{dx}\ln a+\frac{d}{dx}\ln x \\ &\color{#4257b2}\text{Derivative of constant is 0} \\ &=0+\frac{d}{dx}\ln x \\ &=\frac{d}{dx}\ln x \\ &=\frac{d}{dx}\ln |x| \end{align*}

\square

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