Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises: 55

Answer

$y = 1.5x + 0.5$ $y = -\frac{2}{3}x + 2\frac{2}{3}$

Work Step by Step

first of all lets find the derivative of y: $y = x-\sqrt x=x+x^{0.5}$ $y' = (x+x^{0.5})'= (x)'+(x^{0.5})' =1+0.5x^{-0.5}$ So the slope of the tangent line at $(1, 2)$ is (x = 1, and calculate the derivative) $1+0.5\times 1^{-0.5}=1.5$ We use the point-slope form to write an equation of the tangent line at $(1,2)$ $y - 2 = 1.5 (x-1)$ $y = 1.5x - 1.5 + 2 = 1.5x + 0.5$ The slope of the normal line at $(1,2)$ is the negative reciprocal of $1.5$, namely $-\frac{2}{3}$, so an equation is $y - 2 = -\frac{2}{3}(x-1)$ $y = -\frac{2}{3}x +\frac{2}{3} + 2 = -\frac{2}{3}x + 2\frac{2}{3}$
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