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: 21


$$\frac{dy}{dx}=-\frac{x}{y}$$ $$\frac{d^2y}{dx^2}=-\frac{x^2+y^2}{y^3}$$

Work Step by Step

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