## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 152: 8

#### Answer

$$\frac{d}{dx}\sin \frac{y}{x}= \left(\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}\right)\cos\frac{y}{x}.$$

#### Work Step by Step

Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(\sin x)'=\cos x$. Using the quotient and chain rules, we have $$\frac{d}{dx}\sin \frac{y}{x}= \frac{x\frac{dy}{dx}-y}{x^2}\cos\frac{y}{x}= \left(\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}\right)\cos\frac{y}{x}.$$

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