Single Variable Calculus, Chapter 3, 3.9, Section 3.9, Problem 12

a.) Determine the differential of $\displaystyle y = \frac{s}{1 + 2s}$

Using Differential Approximation

$dy = f'(s) ds$



$
\begin{equation}
\begin{aligned}

\frac{dy}{ds} =& \frac{\displaystyle (1 + 2s) \frac{d}{ds} (s) - (s) \frac{d}{ds} (1 + 2s)}{(1 + 2s)^2}
\\
\\
dy =& \left[ \frac{(1 + 2s) (1) - (s)(2)}{(1 + 2s)^2} \right] ds
\\
\\
dy =& \left[ \frac{1 + \cancel{2s}-\cancel{2s}}{(1 + 2s)^2} \right] ds
\\
\\
dy =& \frac{1}{(1 + 2s)^2} ds

\end{aligned}
\end{equation}
$


b.) Determine the differential of $y = u \cos u$

Using Differential Approximation

$dy = f'(u) du$


$
\begin{equation}
\begin{aligned}

\frac{dy}{du} =& (u) \frac{d}{du} (\cos u) + (\cos u) \frac{d}{du} (u)
\\
\\
dy =& [ (u)(- \sin u) + (\cos u) (1) ] du
\\
\\
dy =& -u \sin u + \cos u
\\
\\
dy =& \cos u - u \sin u


\end{aligned}
\end{equation}
$