University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 202: 74



Work Step by Step

$$y^2=2e^{-1/x}$$ Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$2y\times y'=2e^{-1/x}\times\Big(-\frac{1}{x}\Big)'$$ $$2yy'=-2e^{-1/x}\times\frac{(1)'x-1(x)'}{x^2}$$ $$2yy'=-2e^{-1/x}\times\frac{0\times x-1\times1}{x^2}$$ $$2yy'=-2e^{-1/x}\times\frac{-1}{x^2}$$ $$2yy'=\frac{2e^{-1/x}}{x^2}$$ $$yy'=\frac{e^{-1/x}}{x^2}$$ Finally, calculate for $y'$: $$y'=\frac{e^{-1/x}}{x^2y}$$
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