# Thread: Derivative of This Function

1. ## Derivative of This Function

$20[(x/12) - ln(x/12)] +30$

??

I'm thinking...

rearrange $ln(x/12)$ to be
$ln (x) - ln (12)$

so derivative = $20[(1/12) - (1/x) - (1/12)]$

$= -20/x$

??

2. Originally Posted by Equality
$20[(x/12) - ln(x/12)] +30$
You are on the right track

$f(x)=20[(x/12) - ln(x/12)] +30 = \frac{20}{12}x -20ln(x) + 20 ln(12) + 30$

$f'(x)=\frac{20}{12}-\frac{20}{x}$

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

3. thanks for your reply.

so it looks as though where i went wrong is calculating $ln(12)$

i used $1/12$ instead of (derivative of 12)/12 i.e. $0/12$ i.e. $0$

4. 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 $\pi$ is 0. The derivative of $\sqrt2$ is 0. So The derivative of $ln(12)=ln(2^23)=2ln(2)+ln(3) \approx 2(.69314) + 1.09861=2.48489$ is just 0 as well.

Sometimes these sorts of things can cause confusion, but for every $c\in \mathbb{R}$, we have $\frac{d}{dx}c=0$.

5. got it!

thanks heaps Gamma!