# Logarithmic differentiation issue Trying to understand a solution I was given to a problem I was told to use logarithmic differentiation on. 1/x(x+1)(x+2) and I know that log((ab)/c)=log(a)+log(b)−log(c)

Logarithmic differentiation issue
Trying to understand a solution I was given to a problem I was told to use logarithmic differentiation on.
$1/x\left(x+1\right)\left(x+2\right)$
and I know that
$log\left(\left(ab\right)/c\right)=log\left(a\right)+log\left(b\right)-log\left(c\right)$
So I tried to use that rule here and did:
$ln\left(a\right)-ln\left(b\right)-ln\left(c\right)$
and got:
$ln\left(1\right)-ln\left(x\left(x+1\right)-ln\left(x\left(x+2\right)$<>rbwhich simplifies to:
$0-ln\left({x}^{2}+x\right)-ln\left({x}^{2}+2x\right)$
and then I look at the solution which gives:
$y‘=\left(1/\left(x\left(x+1\right)\left(x+2\right)\right)\right)\ast \left(1/x+1/\left(x+1\right)+1/\left(x+2\right)\right)$
I'm just kind of confused on what I am doing wrong.
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Besagnoe9
You have that
$\mathrm{ln}f\left(x\right)=\mathrm{ln}\frac{1}{x\left(x+1\right)\left(x+2\right)}=\mathrm{ln}1-lnx-\mathrm{ln}\left(x+1\right)-\mathrm{ln}\left(x+2\right)$
$=-lnx-\mathrm{ln}\left(x+1\right)-\mathrm{ln}\left(x+2\right).$
Taking derivatives:
$\frac{{f}^{\prime }\left(x\right)}{f\left(x\right)}=-\frac{1}{x}-\frac{1}{x+1}-\frac{1}{x+2}.$
Thus
${f}^{\prime }\left(x\right)=\frac{-1}{x\left(x+1\right)\left(x+2\right)}\left(\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}\right).$