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$g(x) = \displaystyle\frac{x + 1}{(x + 2)^{10}}$

by the quotient rule I get

$ \begin{align} g'(x) & = \displaystyle\frac{1.(x + 2)^{10} - 10.1.(x + 2)^9(x + 1)}{(x + 2)^{20}}\\ & = \displaystyle\frac{x + 2}{(x + 2)^{11}} - \frac{10.(x + 1)}{(x + 2)^{11}} \\ & = \displaystyle\frac{-9x - 8}{(x + 2)^{11}} \end{align} $

my handbook however gives the answer as

$\displaystyle\frac{1 - 9x}{(x + 2)^{11}}$

I just can't see where I made a mistake

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    $\begingroup$ @Joe: $$\frac{(x+2)^{10}}{(x+2)^{20}}=\frac1{(x+2)^{10}}=\frac{x+2}{(x+2)^{11}}$$ $\endgroup$
    – Cameron Buie
    Mar 3, 2014 at 19:13
  • $\begingroup$ Your answer is right. $\endgroup$
    – Woria
    Mar 3, 2014 at 19:16

2 Answers 2

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The answer is this: you did nothing wrong. The handbook is in error. Here's confirmation. Also, there's this.

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    $\begingroup$ Best news I have had all night $\endgroup$
    – Leon
    Mar 3, 2014 at 19:19
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It looks like the handbook is wrong. It seems as if the $10$ was not distributed to $(x+1)$ all the way after differentiating: $$\dfrac{(x+2)^{10}-10(x+2)^9(x+1)}{(x+2)^{20}}=\dfrac{(x+2)^9[(x+2)-10(x+1)]}{(x+2)^{20}}=\dfrac{(-8-9x)}{(x+2)^{11}}$$ Since they did not distribute the $10$ all the way, they got $\dfrac{(1-9x)}{(x+2)^{11}}.$ So you are correct.

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