## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Section 14.3 - Partial Derivatives - Exercises 14.3 - Page 808: 65

#### Answer

$-2$

#### Work Step by Step

In order to find the partial derivative, we will implicitly differentiate with respect to $x$. In doing so, we will treat $y$ as an integer. x + (z^3 + 3z^2x{\partial z}/{\partial x}}) - 2y{\partial z}/{\partial x}} = 0 at (x,y,z) = (1,1,1), it is shown that 1 + ((1)^3 + 3 \times (1)^2 \times (1) \times {\partial z}/{\partial x}} - 2 \times (1) {\partial z}/{\partial x}} = 0 \therefore {\partial z}/{\partial x}} = -2. // previous solution omitted x^2 in the first part of the implicit differentiation.

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