Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 6 - Applications of the Derivative - 6.4 Implicit Differentiation - 6.4 Exercises - Page 334: 27



Work Step by Step

$y^{3}+xy-y= 8x^{4}$ Find the y value: $y^{3}+y-y= 8 $ $y=2$ Calculate by implicit differentiation $3y^{2}\frac{dy}{dx}+x\frac{dy}{dx}+y -\frac{dy}{dx}=32x^{3}$ $(3y^{2}+x-1)\frac{dy}{dx}=32x^{3}-y$ $\frac{dy}{dx}=\frac{32x^{3}-y}{3y^{2}+x-1}$ $m=\frac{5}{2}$ The equation of the tangent line is then found by using the point-slope form of the equation of a line. $y-y_{1}=m(x-x_{1})$ $y-2=\frac{5}{2}(x-1)$ $y=\frac{5}{2}x-\frac{1}{2}$
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