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

Answer

$y=-0.08x+0.54$
1504533593

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.
Small 1504533593
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.