Need to overcome erroneous result when differentiating natural log of a fraction

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 1

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

whose derivative is

$$ \frac{3}{3 x -8 } -\frac{6}{6 x + 2}. $$

Answer 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.

Answer 3

There are two main issues in your question. The first is when you write $\ln(u)=\frac{1}{u}$. This is false (and will usually cost you points on exams). In mathematics, an equality is very precise and specific; if the left hand side is not the same as the right hand side, an equality shouldn't be used.

The second issue is that you're missing the chain rule. The correct statement is $\frac{d}{dx}\ln(u)=\frac{1}{u}\cdot\frac{du}{dx}$. Notice the application of a chain rule (the $\frac{du}{dx}$). Alternately, you might write $(\ln u)'=\frac{u'}{u}$.

Using your setup, you can use $u=\frac{3x-8}{6x+2}$, but don't forget the chain rule! More precisely, $\frac{du}{dx}=\frac{3(6x+2)-6(3x-8)}{(6x+2)^2}=\frac{54}{(6x+2)^2}$. Therefore, $$ \frac{1}{u}\cdot\frac{du}{dx}=\frac{6x+2}{3x-8}\cdot\frac{54}{(6x+2)^2}=\frac{54}{(3x-8)(6x+2)}. $$

Sorry for starting with a rant, but notation errors like that make me upset!