University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 160: 107


See below for proof.

Work Step by Step

$$x^{1/4}=\sqrt{\sqrt x}=(x^{1/2})^{1/2}$$ Using the Chain Rule, we have $$\frac{d}{dx}(x^{1/2})^{1/2}=\frac{1}{2}(x^{1/2})^{-1/2}\frac{d}{dx}(x^{1/2})=\frac{1}{2}(x^{1/2})^{-1/2}\times\frac{1}{2}x^{-1/2}$$ $$=\frac{1}{4}x^{-1/4}\times x^{-1/2}=\frac{1}{4}x^{-3/4}=\frac{1}{4}x^{1/4-1}$$ So the Power Rule $(x^n)'=nx^{n-1}$ holds for the function $x^{1/4}$.
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.