quotient rule difficulties

Question

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.

Answer 1

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.

Answer 2

Your answer is right

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

You can finish it from here....

Answer 3

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

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

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