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consider $$f(x)=\frac{1}{2x-4}$$
The derivative should be $\displaystyle -\frac{1}{2(2x-4)^2}$ However I get $\displaystyle -\frac{2}{(2x-4)^2}$
my workflow: $$\begin{array}{} f'(x)&= &(2x-4)^{-1} \\ &=&-1(2)(2x-4)^{-2} \\ &=&-2(2x-4)^{-2} \end{array}$$
So why does the -2 multiply the denominator and not the numerator? After all, $\displaystyle 2\frac{1}{2}$ is 1 not $\displaystyle \frac{1}{4}$. I feel like I'm missing the obvious.
Thanks all.
On the first glance logging makes things harder, but at the end of the day it doesn't. So, $$ \log f(x) = - \log (2x -4) = - \log 2 - \log (x-2)\\ \frac{d \log f(x)}{dx} = \frac{f'(x)}{f(x)} = -\frac{1}{x-2}\\ f'(x) = -\frac{f(x)}{x-2} =-\frac{1}{2(x-2)^2} $$
The derivative should be $$-\dfrac1{2(x-2)^2}.$$
If so, then your answer is correct since: $$-\dfrac2{(2x-4)^2}=-\dfrac2{(2(x-2))^2}=-\dfrac2{4(x-2)^2}=-\dfrac1{2(x-2)^2}.\tag{$\star$}$$ I will present here 3 different ways to achieve this result:
$$\eqalign{\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{1}{2x-4}\right] &\;=\dfrac{0\cdot(2x-4)-1\cdot 2}{(2x-4)^2} \\ &\;=\dfrac{-2}{(2x-4)^2}\\ &\overset{\displaystyle(\star)}=-\dfrac1{2(x-2)^2}.\\ }$$
Chain rule: $$\dfrac{\mathrm d}{\mathrm dx}f\big(g(x)\big)=f'\big(g(x)\big)\,g'(x).$$
We know that: $$\dfrac{\mathrm d}{\mathrm dx}\dfrac1x=-\dfrac1{x^2}\quad\color{grey}{\text{ and }}\quad\dfrac{\mathrm d}{\mathrm dx}\big(2x-4\big)=2.$$ Therefore by the chain rule: $$\eqalign{ \dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{1}{2x-4}\right] &\;=-\dfrac1{(2x-4)^2}\cdot 2\\ &\;=-\dfrac{2}{(2x-4)^2}\\ &\overset{\displaystyle(\star)}=-\dfrac1{2(x-2)^2}.\\ }$$
This is straightforward: $$\eqalign{\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{1}{2x-4}\right] &\;=\dfrac{-2}{(2x-4)^2}\\ &\overset{\displaystyle(\star)}=-\dfrac1{2(x-2)^2}.\\ }$$
