Single Variable Calculus Questions and Answers

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

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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} $