Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 47


y'=$\frac{2}{\sqrt 4t+9}$

Work Step by Step

The chain rule states that if y=g(h(t)), then y'=g'(h(t))h'(t). Since y=$(4t+9)^{1/2}$, we can set g(t)=$x^{1/2}$ (outside function) and h(t)=4t+9 (inside function), and use the chain rule to find y'. Therefore, y'=g'(4t+9)h'(t) We know that $\frac{d}{dt}$[$t^{1/2}$]=$\frac{1}{2}$$t^{-1/2}$, using the power rule $\frac{d}{dt}$[4t+9]=4, using the power rule Now that we know that g'(t)=$\frac{1}{2}$$t^{-1/2}$ and h'(x)=4, we can find y'. y'=$\frac{1}{2}$$(4t+9)^{-1/2}$(4) =(4)($\frac{1}{2}$)$(4t+9)^{-1/2}$ =$\frac{2}{\sqrt 4t+9}$
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.