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I have: $$\ln(ax + b)^4$$

After calculating ($u'$/$u$) I get:

$$4/(ax+b)$$ but the solution is $$4a/(ax+b)$$

What am I doing wrong? Thanks!

PS: $a$ and $b$ are $\mathbb R^+ $

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You need to apply the chain rule more carefully. Go through all of your steps. –  Qiaochu Yuan Dec 12 '10 at 1:07

1 Answer 1

up vote 1 down vote accepted

You need to apply the chain rule, using $\frac{d}{dx}(ax+b)=a$. (That is your missing factor of $a$.) I can't tell whether you did so, but you can also simplify the first expression to $4\log(ax+b)$.

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Ohhhh! Thanks, I must be tired! –  Qosmo Dec 12 '10 at 1:11

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