Let -2r+5y - z + 3xyz + 163 = 0. Use partial derivatives to calculate and at the point (1, - 3, 2). ax dy Ja,3.2)= %3D ôy 1,3,2)

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Given the equation:

\[ -2x^3 + 5y^3 - z^3 + 3xyz + 163 = 0. \]

Use partial derivatives to calculate \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) at the point (1, -3, 2).

\[
\frac{\partial z}{\partial x}\bigg|_{(1, -3, 2)} = \, \text{\_\_\_\_}
\]

\[
\frac{\partial z}{\partial y}\bigg|_{(1, -3, 2)} = \, \text{\_\_\_\_}
\]
Transcribed Image Text:Given the equation: \[ -2x^3 + 5y^3 - z^3 + 3xyz + 163 = 0. \] Use partial derivatives to calculate \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) at the point (1, -3, 2). \[ \frac{\partial z}{\partial x}\bigg|_{(1, -3, 2)} = \, \text{\_\_\_\_} \] \[ \frac{\partial z}{\partial y}\bigg|_{(1, -3, 2)} = \, \text{\_\_\_\_} \]
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