Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 223: 35

a) $y=\dfrac{1}{x}+C$ b) $y=x+\dfrac{1}{x}+C$ c) $y=5x-\dfrac{1}{x}+C$

a) When $(\dfrac{1}{x})'=-\dfrac{1}{x^2}$ Thus, $y'=-\dfrac{1}{x^2} \implies y=\dfrac{1}{x}+C$ b) When $(x+\dfrac{1}{x})'=1-\dfrac{1}{x^2}$ Thus, $y'=1-\dfrac{1}{x^2} \implies y=x+\dfrac{1}{x}+C$ c) When $(5x-\dfrac{1}{x})'=5+\dfrac{1}{x^2}$ Thus, $y'=5+\dfrac{1}{x^2} \implies y=5x-\dfrac{1}{x}+C$