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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![### Calculus - Partial Derivatives
#### Problem Statement
4. Find the four second partial derivatives for \( f(x,y) = x^2 \sin y \). State final answers and show all work:
\[
\frac{\partial^2 f}{\partial x^2} \quad \quad \frac{\partial^2 f}{\partial y^2}
\]
\[
\frac{\partial^2 f}{\partial x \partial y} \quad \quad \frac{\partial^2 f}{\partial y \partial x}
\]
#### Solution:
Let \( f(x,y) = x^2 \sin y \).
1. **First, find the first partial derivatives:**
- With respect to \( x \):
\[
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (x^2 \sin y) = 2x \sin y
\]
- With respect to \( y \):
\[
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (x^2 \sin y) = x^2 \cos y
\]
2. **Next, find the second partial derivatives:**
- Second partial derivative with respect to \( x \):
\[
\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial}{\partial x} (2x \sin y) = 2 \sin y
\]
- Second partial derivative with respect to \( y \):
\[
\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) = \frac{\partial}{\partial y} (x^2 \cos y) = -x^2 \sin y
\]
- Mixed partial derivative (first \( x \), then \( y \)):
\[
\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F56b9ddaf-a5b9-4606-96af-f9c7010351c4%2F9537e6d6-19a6-49e3-93b7-3b57fa3d7a1d%2Fwyczzkd_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculus - Partial Derivatives
#### Problem Statement
4. Find the four second partial derivatives for \( f(x,y) = x^2 \sin y \). State final answers and show all work:
\[
\frac{\partial^2 f}{\partial x^2} \quad \quad \frac{\partial^2 f}{\partial y^2}
\]
\[
\frac{\partial^2 f}{\partial x \partial y} \quad \quad \frac{\partial^2 f}{\partial y \partial x}
\]
#### Solution:
Let \( f(x,y) = x^2 \sin y \).
1. **First, find the first partial derivatives:**
- With respect to \( x \):
\[
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (x^2 \sin y) = 2x \sin y
\]
- With respect to \( y \):
\[
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (x^2 \sin y) = x^2 \cos y
\]
2. **Next, find the second partial derivatives:**
- Second partial derivative with respect to \( x \):
\[
\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial}{\partial x} (2x \sin y) = 2 \sin y
\]
- Second partial derivative with respect to \( y \):
\[
\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) = \frac{\partial}{\partial y} (x^2 \cos y) = -x^2 \sin y
\]
- Mixed partial derivative (first \( x \), then \( y \)):
\[
\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right
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