Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.4 Tangent Planes and Linear Approximations - 14.4 Exercises - Page 975: 26


$du=\frac{x}{\sqrt{x^2+3y^2}} dx + \frac{3y}{\sqrt{x^2+3y^2}} dy$

Work Step by Step

The differential is defined as: $dw=\frac{\partial w}{\partial x} dx + \frac{\partial w}{\partial y} dy + \frac{\partial w}{\partial z} dz$ Given the function $u=\sqrt{x^2+3y^2}$ . Thus, we find the partial derivatives w.r.t $x$ and $y$: $f_x=\frac{1}{2}(x^2+3y^2)^{-\frac{1}{2}} \cdot 2x=\frac{x}{\sqrt{x^2+3y^2}}$ $f_y=\frac{3y}{\sqrt{x^2+3y^2}}$ Thus, the differential: $du=\frac{x}{\sqrt{x^2+3y^2}} dx + \frac{3y}{\sqrt{x^2+3y^2}} dy$
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