## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 3: Derivatives - Section 3.7 - Implicit Differentiation - Exercises 3.7 - Page 155: 18

#### Answer

$$\frac{{dr}}{{d\theta }} = -\frac{{r + {{\csc }^2}\theta }}{{\sin r + \theta }}$$

#### Work Step by Step

\eqalign{ & \cos r + \cot \theta = r\theta \cr & {\text{differentiate both sides with respect to }}\theta \cr & \frac{d}{{d\theta }}\left( {\cos r} \right) + \frac{d}{{d\theta }}\left( {\cot \theta } \right) = \frac{d}{{d\theta }}\left( {r\theta } \right) \cr & \cr & {\text{use the product rule}} \cr & \frac{d}{{d\theta }}\left( {\cos r} \right) + \frac{d}{{d\theta }}\left( {\cot \theta } \right) = r\frac{d}{{d\theta }}\left( \theta \right) + \theta \frac{d}{{d\theta }}\left( r \right) \cr & \cr & {\text{find derivatives}} \cr & - \sin r\frac{{dr}}{{d\theta }} - {\csc ^2}\theta = r + \theta \frac{{dr}}{{d\theta }} \cr & {\text{Solve for }}\frac{{dr}}{{d\theta }} \cr & -\sin r\frac{{dr}}{{d\theta }} - \theta \frac{{dr}}{{d\theta }} = r + {\csc ^2}\theta \cr & \left( {-\sin r - \theta } \right)\frac{{dr}}{{d\theta }} = r + {\csc ^2}\theta \cr & \frac{{dr}}{{d\theta }} = \frac{{r + {{\csc }^2}\theta }}{{-\sin r - \theta }} \cr}

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