Calculus (3rd Edition)

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

Chapter 15 - Differentiation in Several Variables - 15.4 Differentiability and Tangent Planes - Exercises - Page 789: 4



Work Step by Step

Given $$f(x, y)= \frac{x}{\sqrt{y}} \quad \text { at } \quad(4,4)$$ Since \begin{aligned} f_{x}(x, y) &=\frac{1}{\sqrt{y}} \\ f_{x}(4,4) &= \frac{1}{2} \end{aligned} and \begin{aligned} f_{y}(x, y) &=\frac{-x}{2\sqrt{y^3}} \\ f_{y}(4,4) &= \frac{-1}{4} \end{aligned} Then the tangent plane given by \begin{aligned} z &=f(4,4)+f_{x}(4,4)(x+1)+f_{y}(4,4)(y-2) \\ &=2+\frac{1}{2}(x-4)-\frac{1}{4}(y-4) \end{aligned}
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