Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 154: 55


(a) $y' =\frac{x}{y^2}$ (b) $y''=\frac{y^{3}-2 x^{2}}{y^{5}}$

Work Step by Step

We are given: $$y^{3}-\frac{3}{2} x^{2}=1$$ (a) Differentiate both sides \begin{align*} 3 y^{2} y^{\prime}-3 x&=0\\ y'&=\frac{x}{y^2} \end{align*} and \begin{align*} y''&= \frac{y^2-2xyy'}{y^4} \end{align*} (b) Since $y'=x/y^2$, then \begin{align*} y''&= \frac{y^2-2xyy'}{y^4}\\ &= \frac{y^2-2xy(x/y^2)}{y^4}\\ &= \frac{y^{3}-2 x^{2}}{y^{5}} \end{align*}
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