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: 84


Differentiate implicitly the given equation in both a) and b) to prove the given results.

Work Step by Step

a) $$x^2-y^2=1$$ To show that $dy/dx=x/y$, we will use implicit differentiation to find $dy/dx$ directly. Differentiate both sides of the equation with respect to $x$: $$2x-2y\frac{dy}{dx}=0$$ $$2y\frac{dy}{dx}=2x$$ $$\frac{dy}{dx}=\frac{2x}{2y}=\frac{x}{y}$$ As we can see, $dy/dx$ equals $x/y$ here. b) Next, to show that $d^2y/dx=-1/y^3$, we use implicit differentiation again on $dy/dx$ to find $d^2y/dx^2$ directly: $$\frac{d^2y}{dx^2}=\frac{\frac{dx}{dx}\times y-\frac{dy}{dx}\times x}{y^2}=\frac{y-x\frac{dy}{dx}}{y^2}$$ Here, we replace $dy/dx=x/y$: $$\frac{d^2y}{dx^2}=\frac{y-x\times\frac{x}{y}}{y^2}=\frac{y-\frac{x^2}{y}}{y^2}$$ $$\frac{d^2y}{dx^2}=\frac{\frac{y^2-x^2}{y}}{y^2}=\frac{y^2-x^2}{y^3}$$ From the exercise, we have the equation $x^2y-y^2=1$, which means $y^2-x^2=-1$ $$\frac{d^2y}{dx^2}=\frac{-1}{y^3}$$ As we can see again, $d^2y/dx^2$ equals $-1/y^3$ here.
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