Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 12 - Functions of Several Veriables - 12.5 The Chain Rule - 12.5 Exercises - Page 913: 33

Answer

$\dfrac{dy}{dx}=-\dfrac{y}{x}$

Work Step by Step

Apply Implicit Differentiation Theorem: It states that for a function $F$ to be differentiable on its domain when it satisfies the following condition such that $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ and $F(x,y)=0$ defines $y$ as a differentiable function of $x$ and also, $F_y \ne 0$. We are given that $F(x,y)=2 \sin (xy)-1$ Now, we have: $\dfrac{dy}{dx}=-\dfrac{2y \cos (xy)}{2x \cos (xy)}$ or, $\dfrac{dy}{dx}=-\dfrac{y}{x}$
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.