Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 27

Determine the derivative of the function $\displaystyle y = \frac{r}{\sqrt{r^2 + 1}}$


$\begin{equation} \begin{aligned} y' &= \frac{d}{dr} \left( \frac{r}{r^2+1}\right)\\ \\ y' &= \frac{\left[ (r^2 + 1)^{\frac{1}{2}} \frac{d}{dr} (r)\right] - \left[ (r) \frac{d}{dr} (r^2+1)^{\frac{1}{2}} \right]}{\left[ (r^2+1)^{\frac{1}{2}}\right]^2}\\ \\ y' &= \frac{\left[ (r^2+1)^{\frac{1}{2}}(1)\right]-\left[ (r) \cdot \frac{1}{2} (r^2+1)^{\frac{1}{2}} \frac{d}{dr} (r^2+1) \right]}{r^2+1}\\ \\ y' &= \frac{(r^2+1)^{\frac{1}{2}}- \left( \frac{r}{\cancel{2}} \right) (r^2+1)^{\frac{-1}{2}}(\cancel{2}r)}{r^2+1}\\ \\ y' &= \frac{(r^2+1)^{\frac{1}{2}}- (r^2) (r^2+1)^{\frac{-1}{2}}}{r^2+1}\\ \\ y' &= \frac{(r^2+1)^{\frac{1}{2}} - \frac{r^2}{(r^2+1)^{\frac{1}{2}}}}{r^2+1}\\ \\ y' &= \frac{\cancel{r^2}+1 - \cancel{r^2}}{(r^2+1)(r^2+1)^{\frac{1}{2}}}\\ \\ y' &= \frac{1}{(r^2+1)^{\frac{3}{2}}} \end{aligned} \end{equation}$