Single Variable Calculus, Chapter 3, 3.9, Section 3.9, Problem 16

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

Using Differential Approximation

$dy = f'(x) dx$


$
\begin{equation}
\begin{aligned}

\frac{dy}{dx} =& \frac{d}{dx} \left( \frac{1}{x + 1} \right)
\\
\\
dy =& \left[ \frac{(x + 1) \displaystyle \frac{d}{dx} (1) - (1) \frac{d}{dx} ( x + 1) }{(x + 1)^2} \right] dx
\\
\\
dy =& \left[ \frac{(x + 1) (0) - (1)(1)}{(x + 1)^2} \right] dx
\\
\\
dy =& \frac{-1}{(x + 1)^2} dx

\end{aligned}
\end{equation}
$


b.) Find $dy$ for $x = 1$ and $\displaystyle dx = -0.01$


$
\begin{equation}
\begin{aligned}

dy =& \left[ \frac{-1}{(1 + 1)^2} \right] (-0.01)
\\
\\
dy =& \left[ \frac{-1}{(2)^2} \right] (-0.01)
\\
\\
dy =& \frac{0.01}{4}
\\
\\
dy =& \frac{1}{400} \text{ or } 0.0025

\end{aligned}
\end{equation}
$