Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - Review Exercises - Page 233: 38

Answer

$$y'\left( x \right) = \cot \left( {y - 1} \right)\cot x$$

Work Step by Step

$$\eqalign{ & \sin x\cos \left( {y - 1} \right) = \frac{1}{2} \cr & {\text{use implicit differentiation differentiate both sides with respect to }}x \cr & \frac{d}{{dx}}\left[ {\sin x\cos \left( {y - 1} \right)} \right] = \frac{d}{{dx}}\left[ {\frac{1}{2}} \right] \cr & {\text{use product rule in the left side}} \cr & \sin x\frac{d}{{dx}}\left[ {\cos \left( {y - 1} \right)} \right] + \cos \left( {y - 1} \right)\frac{d}{{dx}}\left[ {\sin x} \right] = \frac{d}{{dx}}\left[ {\frac{1}{2}} \right] \cr & {\text{solving derivatives}} \cr & \sin x\left[ { - \sin \left( {y - 1} \right)} \right]y'\left( x \right) + \cos \left( {y - 1} \right)\left( {\cos x} \right) = 0 \cr & {\text{solving the equation for }}y'\left( x \right) \cr & - \sin x\sin \left( {y - 1} \right)y'\left( x \right) = - \cos \left( {y - 1} \right)\left( {\cos x} \right) \cr & y'\left( x \right) = \frac{{\cos \left( {y - 1} \right)\left( {\cos x} \right)}}{{\sin \left( {y - 1} \right)\sin x}} \cr & {\text{using trigonometric identities}} \cr & y'\left( x \right) = \cot \left( {y - 1} \right)\cot x \cr} $$
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.