Higher Partial Derivative z=2x²-3xy +4y²; d²z 1. oxoy² 3. z+x³ − ln(y³ +x³)= 0; - a²z axoy 0³ z ex²dy 2. 4. U= z = x cosy-ycos.x;- x³ d'u yz' êx³0x² oy
Higher Partial Derivative z=2x²-3xy +4y²; d²z 1. oxoy² 3. z+x³ − ln(y³ +x³)= 0; - a²z axoy 0³ z ex²dy 2. 4. U= z = x cosy-ycos.x;- x³ d'u yz' êx³0x² oy
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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Question
Higher Partial Derivative
![Higher Partial Derivative
0² z
z=2x²-3xy+4y²;-
1.
oxdy²
3. z+x³ −ln(y³ + x³ )= 0;-
a²z
exdy
0³z
ox² dy
2. z=xcosy-ycosx;
x³y²
x ³
r³
d'u
U =
z²
yz' Ox³0x² oy
4.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fccfd7176-9fe5-46a9-8d9a-34e2eda6d8ce%2Feaecb4d4-164d-41a7-bc5a-29cdbcc58676%2Fz9ts7q_processed.png&w=3840&q=75)
Transcribed Image Text:Higher Partial Derivative
0² z
z=2x²-3xy+4y²;-
1.
oxdy²
3. z+x³ −ln(y³ + x³ )= 0;-
a²z
exdy
0³z
ox² dy
2. z=xcosy-ycosx;
x³y²
x ³
r³
d'u
U =
z²
yz' Ox³0x² oy
4.
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Follow-up Questions
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Follow-up Question
find the higher partial derivative of the following
![u
d'u
3
cz³ Ox² Oy
2
x³y² X
.3
+
2
Z
yz](https://content.bartleby.com/qna-images/question/ccfd7176-9fe5-46a9-8d9a-34e2eda6d8ce/8707f3fe-3f0a-4b57-8d1e-a846dd5db055/g7aam94_thumbnail.png)
Transcribed Image Text:u
d'u
3
cz³ Ox² Oy
2
x³y² X
.3
+
2
Z
yz
Solution
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