## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Practice Exercises - Page 202: 75

#### Answer

$$y'=\frac{y}{x}$$

#### Work Step by Step

$$\ln(x/y)=1$$ Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$(\ln(x/y))'=0$$ Recall that $(\ln x)'=1/x$: $$\frac{1}{\frac{x}{y}}\times\Big(\frac{x}{y}\Big)'=0$$ $$\frac{y}{x}\times\frac{x'y-xy'}{y^2}=0$$ $$\frac{y}{x}\times\frac{y-xy'}{y^2}=0$$ $$\frac{y-xy'}{xy}=0$$ Next, we need to separate elements with $y'$ and those without $y'$ into 2 sides of the equation: $$\frac{y}{xy}-\frac{xy'}{xy}=0$$ $$\frac{1}{x}-\frac{y'}{y}=0$$ $$\frac{y'}{y}=\frac{1}{x}$$ Finally, calculate for $y'$: $$y'=\frac{1}{x}\times y=\frac{y}{x}$$

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