# Math Help - Differentiating log of a rational function

1. ## Differentiating log of a rational function

Differentiate $y=log \frac{x+1}{x+4}$

2. Deleted:

3. Wait. I think I got it

Is it:

$y=\log x+1 - \log x+4$
$= \frac{1}{x+1} - \frac{1}{x+4}$
$= 5$

??

4. Originally Posted by nerdzor
Wait. I think I got it

Is it:

$y=\log x+1 - \log x+4$
$= \frac{1}{x+1} - \frac{1}{x+4}$
$= 5$

??
1. Is $\log$ meant to be the same as $\ln$ ?

If so

$y=\log (x+1) - \log( x+4) ~\implies~ y'=\dfrac1{x+1} - \dfrac1{x+4} =$ $\dfrac{(x+4)-(x+1)}{(x+4)(x+1)}= \dfrac3{(x+4)(x+1)}$

2. If log means $\log_{10}$ then you have transform these logarithms into natural logarithms using the base change formula.