## Calculus 10th Edition

Published by Brooks Cole

# Chapter 2 - Differentiation - 2.4 Exercises: 65

#### Answer

$y'=\dfrac{x+4}{\sqrt{x^2+8x}}.$ The derivative at the point $(1, 3)$ is $\dfrac{5}{3}.$

#### Work Step by Step

Using the Chain Rule: $u=x^2+8x$; $\dfrac{du}{dx}=2(x+4)$ $y=\sqrt{u};\dfrac{dy}{du}=\dfrac{1}{2\sqrt{u}}$ $\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=\dfrac{x+4}{\sqrt{x^2+8x}}.$ To evaluate the derivative, plug in $x=1\rightarrow\dfrac{1+4}{\sqrt{1+8}}=\dfrac{5}{3}.$ A graphing utility was used to verify this result.

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.