quotient rule difficulties

I'm trying to use the quotient rule to differentiate $\frac{r}{\sqrt{r^2+1}}$ but I'm getting the wrong answer. So far I have

\begin{align*} \frac {d}{dr} \frac{r}{\sqrt{r^2+1}} &= \frac {\sqrt{r^2+1} \frac {d}{dr} r - r \frac {d}{dr} \sqrt{r^2+1}} {(\sqrt{r^2+1})^2} \\\\\\\\ &= \frac {\sqrt{r^2+1} - r \frac {d}{dr} \sqrt{r^2+1}} {r^2+1} \\\\\\\\ &= \frac {\sqrt{r^2+1} - r \frac{1}{2}(r^2+1)^{-1/2}2r} {(r^2+1)}\\\\\\\\ &= \frac {\sqrt{r^2+1} - r^2 (r^2+1)^{-1/2}} {(r^2+1)}\\\\\\\\ &= (r^2+1)^{-1/2} - r^2 (r^2+1)^{-3/2} \end{align*}

However, the correct answer is just $$(r^2+1)^{-3/2}$$

I've been over it a number of times now, but I can't see the error. I'm pretty new to the quotient rule.

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One way you can guess if two expressions (like your answer and the "correct" answer in this case) are likely be the same function is to plug in a bunch of numbers. If you get the same values for both, then you could try to show that the two are the same function. If you don't, then you KNOW for sure that the two are different. – sibilant Nov 1 '11 at 3:52

$$(r^2+1)^{-1/2} - r^2 (r^2+1)^{-3/2}= \frac{1}{\sqrt{r^2+1}} -\frac{r^2}{(r^2+1)^{3/2}} = \frac{r^2+1}{(r^2+1)^{3/2}} -\frac{r^2}{(r^2+1)^{3/2}}$$

You can finish it from here....

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There's nothing wrong with your application of the quotient rule. You just need to simplify your answer further: $$\begin{eqnarray*} (r^2+1)^{-1/2} - r^2 (r^2+1)^{-3/2} &=& (r^2+1) \cdot (r^2+1)^{-3/2} - r^2 \cdot (r^2+1)^{-3/2} \\ &=& (r^2+1)^{-3/2} \cdot \left((r^2+1) - r^2 \right) \\ &=& (r^2+1)^{-3/2} \cdot 1 \\ &=& (r^2+1)^{-3/2}, \end{eqnarray*}$$ which is the official answer.

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You are all right, just only one step away from the finish line.

$$(r^2+1)^{-1/2} - r^2 (r^2+1)^{-3/2}$$

$$= \dfrac{1}{\sqrt {r^2+1}} - \dfrac {r^2}{ (r^2+1)^{3/2}}= \dfrac{(r^2+1)}{ (r^2+1)^{3/2}} - \dfrac {r^2}{ (r^2+1)^{3/2}}=\dfrac {1}{ (r^2+1)^{3/2}}.$$

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