Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.5 Exercises: 49

Answer

$\frac{d^2y}{dx^2}=\frac{12xy^2-9x^4}{4y^3}$

Work Step by Step

$y^2=x^3$ $\frac{d}{dx}[y^2]=\frac{d}{dx}[x^3]$ $2y\frac{dy}{dx}=3x^2$ $\frac{dy}{dx}=\frac{3x^2}{2y}$ $\frac{d^2y}{dx^2}=\frac{d}{dx}[\frac{dy}{dx}]$ $\frac{d^2y}{dx^2}=\frac{d}{dx}[\frac{3x^2}{2y}]$ $\frac{d^2y}{dx^2}=\frac{6x2y-3x^22\frac{dy}{dx}}{(2y)^2}$ $\frac{d^2y}{dx^2}=\frac{12xy-6x^2\frac{dy}{dx}}{4y^2}$ $\frac{d^2y}{dx^2}=\frac{12xy-6x^2\frac{3x^2}{2y}}{4y^2}$ $\frac{d^2y}{dx^2}=\frac{12xy-\frac{18x^4}{2y}}{4y^2}$ $\frac{d^2y}{dx^2}=\frac{12xy-\frac{18x^4}{2y}}{4y^2} \times \frac{2y}{2y}$ $\frac{d^2y}{dx^2}=\frac{24xy^2-18x^4}{8y^3}$ $\frac{d^2y}{dx^2}=\frac{12xy^2-9x^4}{4y^3}$
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