<s>[INST] 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.
 [/INST] 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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