Answer
$y'=\frac{1}{\sqrt{-x^2-x}}$
Work Step by Step
Apply the Chain Rule:
$y=sin^{-1}(2x+1)\\
y'=\frac{1}{\sqrt{1-(2x+1)^2}}.2\\
y'=\frac{1}{\sqrt{1-4x^2-4x-1}}.2\\
y'=\frac{1}{\sqrt{-4x^2-4x}}.2\\
y'=\frac{1}{\sqrt{-x^2-x}}$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 216: 51
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