$\displaystyle 20[(x/12) - ln(x/12)] +30$
??
I'm thinking...
rearrange $\displaystyle ln(x/12)$ to be
$\displaystyle ln (x) - ln (12)$
so derivative = $\displaystyle 20[(1/12) - (1/x) - (1/12)]$
$\displaystyle = -20/x$
??
You are on the right track
$\displaystyle f(x)=20[(x/12) - ln(x/12)] +30 = \frac{20}{12}x -20ln(x) + 20 ln(12) + 30$
$\displaystyle f'(x)=\frac{20}{12}-\frac{20}{x}$
All those terms without x's in them are just real numbers, so their derivative is 0, so there are really only two terms you have to worry about
thanks for your reply.
so it looks as though where i went wrong is calculating $\displaystyle ln(12)$
i used $\displaystyle 1/12$ instead of (derivative of 12)/12 i.e. $\displaystyle 0/12$ i.e. $\displaystyle 0$
Yeah, I guess that is one way of thinking about it.
Alternatively you know ln(12) is just a number you know? Like the derivative of 3 is 0. The derivative of $\displaystyle \pi$ is 0. The derivative of $\displaystyle \sqrt2$ is 0. So The derivative of $\displaystyle ln(12)=ln(2^23)=2ln(2)+ln(3) \approx 2(.69314) + 1.09861=2.48489$ is just 0 as well.
Sometimes these sorts of things can cause confusion, but for every $\displaystyle c\in \mathbb{R}$, we have $\displaystyle \frac{d}{dx}c=0$.