Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.2 Exercises: 2

Answer

$m_{a}=-\frac{1}{2}$ $m_{b}= -1$

Work Step by Step

The slope of the tangent line at a given point is equivalent to the derivative of the line evaluated at that point. $a.) y = x^{-\frac{1}{2}}$ Use the power rule to differentiate: $y' = -\frac{1}{2}x^{-\frac{3}{2}}$ Now since the point we are evaluating is at one, 1 to any exponent is itself. Leaving us with $-\frac{1}{2}$ as the slope of the tangent line at $(1,1)$. $y' = -\frac{1}{2}1^{-\frac{3}{2}}$ $y' = -\frac{1}{2}$ Now do the same thing with the other function. $b.) y = x^{-1}$ $y' = -1x^{-2}$ $y' = -1 (1)^{-2}$ $y' = -1$
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