Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises - Page 188: 36



Work Step by Step

Let's start off by finding the derivative of the original function $f(x)=\frac{x}{1+x^2}$ We're going to have to use the quotient rule, the quotient rule states that: $\frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$ So using that, we can now find the derivative of $f(x)$. Let's start by finding the derivative of the numerator and denominator: Numerator: $x$ Derivative of Numerator: $1$ Denominator: $1+x^2$ Derivative of Denominator: $2x$ So now let's just plug that into the quotient rule: $f'(x)=\frac{(1)(1+x^2)-(2x)(x)}{(1+x^2)^2}=\frac{1+x^2-2x^2}{(1+x^2)^2}=\frac{1-x^2}{(1+x^2)^2}$ Now, we need to find the slope of the tangent line at $(3,0.3)$. Let's plug in $x=3$: $f'(3)=\frac{1-(3)^2}{(1+(3)^2)^2}=\frac{1-9}{(1+9)^2}=\frac{-8}{100}=-0.08$ So now we know that the slope $m=-0.08$ at the point $(3,0.3)$. We can use the point-slope form to create the equation of a line. $y-y_1=m(x-x_1)$ Now we can plug in our information: $y-0.3=-0.08(x-3)$ $y-0.3=-0.08x+0.24$ $y=-0.08x+0.54$ And that's your final answer. Now you can graph both the original function and the tangent line on the same graph to verify your answer.
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