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]
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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! $\displaystyle \frac{d}{\,dx}\log_2(1-3x)=\frac{1}{(1-3x)\ln 2}\cdot -3=\frac{-3}{(1-3x)\ln 2}=\color{red}\boxed{\frac{3}{(3x-1)\ln 2}}$ Does this make sense? --Chris
yes, it does! i completely forgot!, thanks
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