Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 88

Differentiate $\displaystyle w = \frac{u}{\sqrt{1 + u^2}}$

By applying Quotient Rule and Chain Rule, we get

$
\begin{equation}
\begin{aligned}
w'(u) &= \frac{(1 + u^2)^{\frac{1}{2}} \cdot \frac{d}{du} (u) - (u) \cdot \frac{d}{du} (1 + u^2)^{\frac{1}{2}} }{\left[(1 + u^2)^{\frac{1}{2}} \right]^2}\\
\\
w'(u) &= \frac{(1 + u^2)^{\frac{1}{2}}(1) - (u) \cdot \frac{1}{2} (1 + u^2)^{\frac{1}{2}-1} \cdot \frac{d}{du} ( 1 + u^2) }{1 + u^2}\\
\\
w'(u) &= \frac{(1 + u^2)^{\frac{1}{2}} - u \cdot \frac{1}{2} ( 1 + u^2)^{-\frac{1}{2}} (2u) }{1 + u^2}\\
\\
w'(u) &= \frac{(1 + u^2)^{\frac{1}{2}} - \frac{u^2}{(1 + u^2)^{\frac{1}{2}}} }{1 + u^2}\\
\\
w'(u) &= \frac{1 + u^2 - u^2}{(1 + u^2)^{\frac{1}{2}} (1 + u^2) }\\
\\
&= \frac{1}{(1 + u^2)^{\frac{3}{2}}}
\end{aligned}
\end{equation}
$