Let f be a function with continuous second partial derivatives and define: z = f (2x³y - y₁y³) if you take u = 2x³y - y and w = y³ is obtained: (22³1) fu + 5y¹.fw From the above it is concluded that: อะ A) Zyx = 30x²y5. fwu + (12r5y − 6x²y) · fuu + 6x² · fu - . ду B) Zyx = 6x²y (2x³− 1). fuu + 6x² fu + 6x²y. fwu D) Zyx = 6x² fu 6x²y + 5y4. fwu 6x²y . = . - C) Zyx = 6x². fu+ (2x³ − 1) · fuu + 5y¹. fwu · 6x²y .

Calculus: Early Transcendentals
8th Edition
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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Let f be a function with continuous second partial derivatives and define:
= f (2x³y – y, y³)
if you take u = 2x³y - y and
From the above it is concluded that:
B) Zyx
2=
=
A) Zyx = 30x²y5 fwu + (12x5y - 6x²y). fuu + 6x². fu
.
əz
w = y³ is obtained: ду
D) Zyx =
=
= 6x² fu 6x²y + 5y4. fwu 6x²y
(231) fu + 5y4. fw
6x²y (2x³ - 1) fuu + 6x². fu + 6x²y.fwu
.
C) * = = 6x² · fu + (2x³ — 1) · fuu + 5y¹ · fwu 6x²y
.
Transcribed Image Text:Let f be a function with continuous second partial derivatives and define: = f (2x³y – y, y³) if you take u = 2x³y - y and From the above it is concluded that: B) Zyx 2= = A) Zyx = 30x²y5 fwu + (12x5y - 6x²y). fuu + 6x². fu . əz w = y³ is obtained: ду D) Zyx = = = 6x² fu 6x²y + 5y4. fwu 6x²y (231) fu + 5y4. fw 6x²y (2x³ - 1) fuu + 6x². fu + 6x²y.fwu . C) * = = 6x² · fu + (2x³ — 1) · fuu + 5y¹ · fwu 6x²y .
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