Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.4 Exercises: 27

Answer

$y'=\dfrac{1}{\sqrt{(x^2+1)^3}}.$

Work Step by Step

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=x; u'(x)=1$ $v(x)=\sqrt{x^2+1}$ $v'(x)$ is found using the Chain Rule: $u=(x^2+1)$; $\dfrac{du}{dx}=2x$ $\dfrac{d}{du}v(u)=\dfrac{1}{2\sqrt{u}}$ $\dfrac{d}{dx}v(x)=\dfrac{d}{du}v(u)\times\dfrac{du}{dx}=(2x)\dfrac{1}{2\sqrt{x^2+1}}=\dfrac{x}{\sqrt{x^2+1}}.$ $y'=\dfrac{(1)(\sqrt{x^2+1})-(\dfrac{x}{\sqrt{x^2+1}})(x)}{(\sqrt{x^2+1})^2}$ $=\dfrac{(\dfrac{1}{\sqrt{x^2+1}})(x^2+1-x^2)}{x^2+1}$ $=\dfrac{1}{\sqrt{(x^2+1)^3}}.$
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.