Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.7 - Implicit Differentiation - Exercises 3.7 - Page 155: 43

Answer

point $(-3,2)$, $m_1=\frac{-27}{8}$ point $(3,2)$, $m_2=\frac{27}{8}$ point $(-3,-2)$, $m_3=\frac{27}{8}$ point $(3,-2)$, $m_4=\frac{27}{-8}$

Work Step by Step

Step 1. Given the equation $y^4-4y^2=x^4-9x^2$, differentiate both sides with respect to $x$: $4y^3y'-8yy'=4x^3-18x$ which gives $y'=\frac{2x^3-9x}{2y^3-4y}$, Step 2. the slope of the tangent line at point $(-3,2)$ is $m_1=y'=\frac{2(-3)^3-9(-3)}{2(2)^3-4(2)}=\frac{-54+27}{16-8}=\frac{-27}{8}$ Step 3. the slope of the tangent line at point $(3,2)$ is $m_2=y'=\frac{2(3)^3-9(3)}{2(2)^3-4(2)}=\frac{54-27}{16-8}=\frac{27}{8}$ Step 4. the slope of the tangent line at point $(-3,-2)$ is $m_3=y'=\frac{2(-3)^3-9(-3)}{2(-2)^3-4(-2)}=\frac{-54+27}{-16+8}=\frac{27}{8}$ Step 5. the slope of the tangent line at point $(3,-2)$ is $m_4=y'=\frac{2(3)^3-9(3)}{2(-2)^3-4(-2)}=\frac{54-27}{-16+8}=\frac{27}{-8}$
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