Calculus 10th Edition

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

Chapter 2 - Differentiation - 2.4 Exercises - Page 137: 82

Answer

$y=2x-1.$

Work Step by Step

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=|x| ; u'(x)=\dfrac{x}{|x|}$ $v(x)=\sqrt{2-x^2} ; v'(x)=-\dfrac{x}{\sqrt{2-x^2}}$ $f'(x)=\dfrac{(\dfrac{x}{|x|})(\sqrt{2-x^2})-(|x|)(-\dfrac{x}{\sqrt{2-x^2}})}{2-x^2}$ $=\dfrac{2x}{|x|\sqrt{(2-x^2)^3}}$ $f'(1)=\dfrac{2(1)}{(1)\sqrt{(2-1^2)^3}}=2.$ Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$. $(y-1)=2(x-1)\rightarrow y=2x-1.$ A graphing utility is used to graph the function f and the specified tangent line. The intersection point at (1, 1) is confirmed.
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