# Math Help - derivatives of logarithmic functions

1. ## derivatives of logarithmic functions

I got a different answer from the book, and I still right? what happened here?

differentiate

f(x)=log 2 (1-3x)

Book answer: f'(x)=3/[(3x-1) ln 2]

My answer: f'(x)=1/[(1-3x) ln 2]

2. Originally Posted by thecount
I got a different answer from the book, and I still right? what happened here?

differentiate

f(x)=log 2 (1-3x)

Book answer: f'(x)=3/[(3x-1) ln 2]

My answer: f'(x)=1/[(1-3x) ln 2]

You forgot to apply the chain rule!

$\frac{d}{\,dx}\log_2(1-3x)=\frac{1}{(1-3x)\ln 2}\cdot -3=\frac{-3}{(1-3x)\ln 2}=\color{red}\boxed{\frac{3}{(3x-1)\ln 2}}$

Does this make sense?

--Chris

3. yes, it does! i completely forgot!, thanks