University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises - Page 136: 57


The tangent to $f(x)$ at the origin is $y=4x$. The tangent to $f(x)$ at $(1,2)$ is $y=2$

Work Step by Step

$$y=f(x)=\frac{4x}{x^2+1}$$ 1) First, we look for $f'(x)$, applying the Derivative Quotient Rule: $$f'(x)=\frac{(4x)'(x^2+1)-4x(x^2+1)'}{(x^2+1)^2}=\frac{4(x^2+1)-4x\times2x}{(x^2+1)^2}$$ $$f'(x)=\frac{4x^2+4-8x^2}{(x^2+1)^2}=\frac{4-4x^2}{(x^2+1)^2}$$ 2) At the origin: $$f'(0)=\frac{4-4\times0^2}{(0^2+1)^2}=\frac{4-0}{1}=4$$ The tangent to $f(x)$ at the origin is $$y-0=f'(0)(x-0)$$ $$y=f'(0)x=4x$$ 3) At $(1,2)$: $$f'(1)=\frac{4-4\times1^2}{(1^2+1)^2}=\frac{4-4}{2^2}=\frac{0}{4}=0$$ The tangent to $f(x)$ at $(1,2)$ is $$y-2=f'(1)(x-1)=0(x-1)$$ $$y-2=0$$ $$y=2$$
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.