University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.7 - Implicit Differentiation - Exercises - Page 164: 27


At $(2,2)$, $$\frac{d^2y}{dx^2}=-2$$

Work Step by Step

$$x^3+y^3=16$$ 1) Find $dy/dx$: Differentiate both sides of the equation with respect to $x$: $$3x^2+3y^2\frac{dy}{dx}=0$$ Isolate the terms with $dy/dx$ into one side and solve for $dy/dx$: $$3y^2\frac{dy}{dx}=-3x^2$$ $$\frac{dy}{dx}=-\frac{x^2}{y^2}$$ 2) Find $d^2y/dx^2$: Differentiate the derivative $dy/dx$ as normal, using Quotient Rule: $$\frac{d^2y}{dx^2}=\frac{d}{dx}\Big(-\frac{x^2}{y^2}\Big)=-\frac{2xy^2-x^2(2y\frac{dy}{dx})}{y^4}=-\frac{2xy^2-2x^2y\frac{dy}{dx}}{y^4}$$ Now we can use the result $dy/dx=-x^2/y^2$ from part 1) to do substitution here: $$\frac{d^2y}{dx^2}=-\frac{2xy^2-2x^2y\Big(-\frac{x^2}{y^2}\Big)}{y^4}$$ $$\frac{d^2y}{dx^2}=-\frac{2xy^2+\frac{2x^4}{y}}{y^4}=-\frac{\frac{2xy^3+2x^4}{y}}{y^4}=-\frac{2xy^3+2x^4}{y^5}$$ - At $(2,2)$: $$\frac{d^2y}{dx^2}=-\frac{2\times2\times2^3+2\times2^4}{2^5}=-\frac{2^5+2^5}{2^5}=-\frac{2^6}{2^5}=-2$$
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