Question: I am trying to differentiate the following:

$$ln(3x-8/6x+2)$$

my (incorrect) method is:

let $$ln(x) = ln(u)$$
therefore when differentiating u..
$$ln(u) = 1/u$$
and diff of$$$$(3x-8/6x+2) = 3/6 = 0.5$$$$
so punching in values gives
$$0.5 \* (1/(3x-8/6x+2)) = (0.5/(3x-8/6x+2)) $$

However I know the answer and it is not what I have derived, I do not know how to get to the answer. it is here <https://archive.uea.ac.uk/jtm/10/lec10p3.pdf>
question h) page 10.

Any help showing where I have gone wrong would be very much appreciated.

Answer: 
$$ ln[(3x-8)/(6x+2)] =ln[3x-8 ]- ln [6x+2 ] $$

whose derivative is 

$$ \frac{3}{3 x -8 } -\frac{6}{6 x + 2}. $$
The derivative of an $f(g(x))$ function, considering that both functions are differentiable, is : $f'(g(x))g'(x)$ from the chain rule.

What you have to find is : $ \frac{d}{dx}f(x) = \frac{d}{dx} ln(\frac{3x-8}{6x+2}) $

Note that :$ [ln(g(x))]' = \frac{1}{g(x)}g'(x) $ . 

Then it is :$ \frac{d}{dx}f(x) = f'(x) = [ln(\frac{3x-8}{6x+2})]' = \frac{6x+2}{3x-8}( \frac{6x+2}{3x-8})' $. 

You can also write $ ln(\frac{3x-8}{6x+2}) = ln(3x-8) - ln(6x+2) $ and then differentiate this expression.

You can continue from then on.