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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![**Problem Statement:**
Given the function \( f(x,y) = x^3 (x^2 y^3 - 2y^2) \), find all of the first and second partial derivatives.
**Solution:**
We start by finding the first partial derivatives of the function \( f \) with respect to \( x \) and \( y \).
**Finding the First Partial Derivative with respect to \( x \) (\( f_x \)):**
Given:
\[ f(x,y) = x^3 (x^2 y^3 - 2y^2) \]
Let's expand the expression first:
\[ f(x,y) = x^5 y^3 - 2 x^3 y^2 \]
Now, take the partial derivative with respect to \( x \):
\[ f_x = \frac{\partial}{\partial x} (x^5 y^3 - 2 x^3 y^2) \]
\[ f_x = 5 x^4 y^3 - 6 x^2 y^2 \]
**Finding the First Partial Derivative with respect to \( y \) (\( f_y \)):**
Again, take the partial derivative with respect to \( y \):
\[ f_y = \frac{\partial}{\partial y} (x^5 y^3 - 2 x^3 y^2) \]
\[ f_y = 3 x^5 y^2 - 4 x^3 y \]
Next, we find the second partial derivatives. We need \( f_{xx}, f_{xy}, f_{yx}, f_{yy} \).
**Finding the Second Partial Derivative with respect to \( x \) (\( f_{xx} \)):**
\[ f_{xx} = \frac{\partial}{\partial x} (5 x^4 y^3 - 6 x^2 y^2) \]
\[ f_{xx} = 20 x^3 y^3 - 12 x y^2 \]
**Finding the Second Partial Derivative with respect to \( y \) (\( f_{yy} \)):**
\[ f_{yy} = \frac{\partial}{\partial y} (3 x^5 y^2 - 4 x^3 y) \]
\[ f_{yy} = 6 x^5 y - 4 x^3 \]
**Finding the Mixed Partial Derivative](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3e8fda94-3d30-45d6-a666-29bafa6b8019%2Fd29122a7-fd1a-4ac8-8b1d-7f802eb1489e%2Fdx3or8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Given the function \( f(x,y) = x^3 (x^2 y^3 - 2y^2) \), find all of the first and second partial derivatives.
**Solution:**
We start by finding the first partial derivatives of the function \( f \) with respect to \( x \) and \( y \).
**Finding the First Partial Derivative with respect to \( x \) (\( f_x \)):**
Given:
\[ f(x,y) = x^3 (x^2 y^3 - 2y^2) \]
Let's expand the expression first:
\[ f(x,y) = x^5 y^3 - 2 x^3 y^2 \]
Now, take the partial derivative with respect to \( x \):
\[ f_x = \frac{\partial}{\partial x} (x^5 y^3 - 2 x^3 y^2) \]
\[ f_x = 5 x^4 y^3 - 6 x^2 y^2 \]
**Finding the First Partial Derivative with respect to \( y \) (\( f_y \)):**
Again, take the partial derivative with respect to \( y \):
\[ f_y = \frac{\partial}{\partial y} (x^5 y^3 - 2 x^3 y^2) \]
\[ f_y = 3 x^5 y^2 - 4 x^3 y \]
Next, we find the second partial derivatives. We need \( f_{xx}, f_{xy}, f_{yx}, f_{yy} \).
**Finding the Second Partial Derivative with respect to \( x \) (\( f_{xx} \)):**
\[ f_{xx} = \frac{\partial}{\partial x} (5 x^4 y^3 - 6 x^2 y^2) \]
\[ f_{xx} = 20 x^3 y^3 - 12 x y^2 \]
**Finding the Second Partial Derivative with respect to \( y \) (\( f_{yy} \)):**
\[ f_{yy} = \frac{\partial}{\partial y} (3 x^5 y^2 - 4 x^3 y) \]
\[ f_{yy} = 6 x^5 y - 4 x^3 \]
**Finding the Mixed Partial Derivative
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