University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.7 - Implicit Differentiation - Exercises - Page 164: 3



Work Step by Step

Apply rule of differentiation: $y'=a'(x)+b'(x)$ Here, $a'(x)=m'(x)n(x)+m(x)n'(x)$ Given: $2xy+y^2=x+\frac{y}{d/dx}$ $2(x'y+xy')+2yy'=1+y'$ $2(y+xy')+2yy'=1+y'$ $2y+2xy'+2yy'=1+y'$ $2xy'+2yy'-y'=1-2y$ $y'(2x+2y-1)=1-2y$ Hence, $y'=\dfrac{1-2y}{2x+2y-1}$
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