## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises: 29

#### Answer

a) $F'(2) = -\frac{3}{5}$ b) The red is the graph of $f(x) = \frac{5x}{1+x^2}$ The blue is the tangent line $y = -\frac{3x}{5} + \frac{16}{5}$.

#### Work Step by Step

$F'(a) = \lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a}$ $F'(a) = \lim\limits_{x \to a} \frac{\frac{5x}{1+x^2}-\frac{5a}{1+a^2}}{(x-a)}$ $F'(a) = \lim\limits_{x \to a} \frac{\frac{5x}{1+x^2}-\frac{5a}{1+a^2}}{(x-a)}$ $\times {\frac{(1+x^2)(1+a^2)}{(1+x^2)(1+a^2}}$ $F'(a) = \lim\limits_{x \to a} \frac{5x(1+a^2)-5a(1+x^2)}{(x-a)(1+x^2)(1+a^2)}$ $F'(a) = \lim\limits_{x \to a} \frac{5x+5xa^2-5a-5x^2)}{(x-a)(1+x^2)(1+a^2)}$ Simplify the equation: $F'(a) = \lim\limits_{x \to a} \frac{5[(xa^2-a)-(x^2a-x)]}{(x-a)(1+x^2)(1+a^2)}$ $F'(a) = \lim\limits_{x \to a} \frac{5[(a(xa-1))-x(xa-1)]}{(x-a)(1+x^2)(1+a^2)}$ $F'(a) = \lim\limits_{x \to a} \frac{5[(a-x)(xa-1)]}{(x-a)(1+x^2)(1+a^2)}$ Multiply $-1$ to $(a-x)$ to make it $(x-a)$ $F'(a) = \lim\limits_{x \to a} \frac{-5[(a-x)(xa-1)]}{(x-a)(1+x^2)(1+a^2)}$ $F'(a) = \lim\limits_{x \to a} \frac{-5[(x-a)(xa-1)]}{(x-a)(1+x^2)(1+a^2)}$ $F'(a) = \lim\limits_{x \to a} \frac{-5[(xa-1)]}{(1+x^2)(1+a^2)}$ Substitute $x$ for $a$ $F'(a) = \lim\limits_{x \to a} \frac{-5(a*a-1)}{(1+a^2)(1+a^2)}$ $F'(a) = \frac{-5(a^2-1)}{(1+a^2)^2}$ $F'(2) = \frac{-5(2^2-1)}{(1+2^2)^2}$ $F'(2) = \frac{-5(4-1)}{(1+4)^2}$ $F'(2) = \frac{-5(3)}{5^2}$ $F'(2) = -\frac{-15}{25}$ $F'(2) = -\frac{3}{5}$ Tangent line: $y - y_{1} = m(x-x_{1})$ $y - 2 = -\frac{3}{5}(x-2)$ $y = -\frac{3}{5}(x-2) + 2$ $y = -\frac{3}{5}x + \frac{6}{5}+2$ $y = -\frac{3}{5}x + \frac{16}{5}$

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