University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 198: 19


$dy=(3x^2-\frac{3}{2\sqrt {x}}){dx}$

Work Step by Step

We evaluate the function $y=x^3-3\sqrt{x}$ on differentiating the above: $\frac{dy}{dx}=\frac{d(x^3-3\sqrt {x})}{dx}$ or $\frac{dy}{dx}=3x^2-\frac{3}{2\sqrt {x}}$ or $dy=(3x^2-\frac{3}{2\sqrt {x}}){dx}$ The final answer is: $dy=(3x^2-\frac{3}{2\sqrt {x}}){dx}$
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