Confused as to how to compute this derivative

Question

The question asks us to compute the derivative of the following:

$\frac{d}{dx}\frac{\left(x+10\right)^{10}}{\left(2x-4\right)^{10}}$

Now, by using the natural log and various rules related to that, I've arrived to:

$f\left(x\right)=10ln\left(x+1\right)-10ln\left(2x-4\right)$

What confuses me is the next step in my professor's notes. He taught this concept to us very quickly so I can't follow what he's doing. He ends up doing the below:

$\frac{1}{f\left(x\right)}\cdot f\:'\left(x\right)=\frac{10}{x+1}\left(1\right)-\frac{10}{2x-4}\left(2\right)$

Which cancels various things out. I don't know why he's multipled the LHS by $\frac{1}{f\left(x\right)}$ or the other terms by (1) and (2). Is he using other log rules for derivatives?

Answer 1

Suppose that $$ f(x)=\frac{(x+10)^{10}}{(2x-4)^{10}}. $$

Then, applying the logarithm to both sides gives $$ \ln f(x)=10\ln(x+10)-10\ln(2x-4). $$

Taking the derivative of both sides gives (remembering the chain rule to give the factor of $2$ in the right-most term). The chain rule also gives the $f'(x)$ on the LHS. $$ \frac{1}{f(x)}\cdot f'(x)=10\cdot \frac{1}{x+10}-10\cdot \frac{1}{2x-4}\cdot 2 $$

Then, multiplying by $f(x)$ on both sides gives $$ f'(x)=\frac{10f(x)}{x+10}-\frac{20f(x)}{2x-4}. $$

Answer 2

$f(x) = ln(g(x))$ where $g(x) = {(x+10)^{10}\over(2x-4)^{10}}$

By the chain rule, $f'(x) = {g'(x)\over g(x)}$

Since $f'(x) = {10\over x+1} + 2{10\over2x-4}$ we get

$$ g'(x) = g(x) \left({10\over x+1} + 2{10\over2x-4} \right) $$