## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 152: 24

#### Answer

$$y'(x) =\frac{\sin y+y \sin x}{\cos x-x \cos y}$$

#### Work Step by Step

Given $$x\sin y - y\cos x=0$$ Differentiate with respect to $x$ \begin{align*} \frac{d}{dx} \left[x\sin y - y\cos x\right]&=0\\ x\cos yy'(x) +\sin y + y\sin x - y'(x)\cos x &=0 \\ \left [x\cos y -\cos x \right] y'(x)& =-\sin y - y\sin x \end{align*} Then $$y'(x) =\frac{\sin y+y \sin x}{\cos x-x \cos y}$$

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