Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises - Page 205: 56

Answer

(a) The equation of the tangent line is: $~~~y = 2x-1$ (b) We can see a sketch of the graphs below.

Work Step by Step

(a) $y = \frac{\vert x \vert}{\sqrt{2-x^2}}$ When $x \geq 1$: $y = \frac{x}{\sqrt{2-x^2}}$ We can find $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{\sqrt{2-x^2}-\frac{1}{2}(2-x^2)^{-1/2}(-2x)(x)}{2-x^2}$ $\frac{dy}{dx} = \frac{\frac{2-x^2}{\sqrt{2-x^2}}+\frac{x^2}{\sqrt{2-x^2}}}{2-x^2}$ $\frac{dy}{dx} = \frac{2}{(2-x^2)^{3/2}}$ When $x = 1~~~$ then $~~\frac{dy}{dx} = \frac{2}{(2-1^2)^{3/2}} = 2$ The slope of the tangent line is $m=2$ We can find the equation of the tangent line: $y-1 = 2(x-1)$ $y = 2x-1$ The equation of the tangent line is: $~~~y = 2x-1$ (b) We can see a sketch of the graphs below.
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