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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Homework: HW 17.2
6 of 13
![**Problem Statement:**
Find \( f_x(x,y) \) and \( f_y(x,y) \). Then, find \( f_x(-1, 4) \) and \( f_y(1, -4) \).
Equation:
\[ f(x,y) = \frac{5x^2 y}{e^{xy} + 3} \]
**Solution:**
Step 1: Find the partial derivative of \( f(x,y) \) with respect to \( x \) (denoted as \( f_x(x,y) \)).
Step 2: Find the partial derivative of \( f(x,y) \) with respect to \( y \) (denoted as \( f_y(x,y) \)).
Step 3: Evaluate \( f_x(x,y) \) at the point \( (-1, 4) \).
Step 4: Evaluate \( f_y(x,y) \) at the point \( (1, -4) \).
**Explanation of Procedures:**
1. **Partial Derivative with Respect to \( x \):**
To differentiate \( f(x,y) = \frac{5x^2 y}{e^{xy} + 3} \) with respect to \( x \), use the quotient rule:
\[
f_x(x,y) = \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}
\]
where \( u = 5x^2 y \) and \( v = e^{xy} + 3 \).
- First, find \( u' \) and \( v' \):
\[ u' = \frac{d}{dx} (5x^2 y) = 10xy \]
\[ v' = \frac{d}{dx} (e^{xy} + 3) = ye^{xy} \]
- Using the quotient rule:
\[
f_x(x,y) = \frac{(10xy)(e^{xy} + 3) - (5x^2 y)(ye^{xy})}{(e^{xy} + 3)^2}
\]
2. **Partial Derivative with Respect to \( y \):**
To differentiate \( f(x,y) \) with respect to \( y \):
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0c1127c9-3767-40bf-bfb8-794124c4dac3%2F71dfb8b1-4952-4e15-9427-142c3f3f112f%2Fakbq6k5_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find \( f_x(x,y) \) and \( f_y(x,y) \). Then, find \( f_x(-1, 4) \) and \( f_y(1, -4) \).
Equation:
\[ f(x,y) = \frac{5x^2 y}{e^{xy} + 3} \]
**Solution:**
Step 1: Find the partial derivative of \( f(x,y) \) with respect to \( x \) (denoted as \( f_x(x,y) \)).
Step 2: Find the partial derivative of \( f(x,y) \) with respect to \( y \) (denoted as \( f_y(x,y) \)).
Step 3: Evaluate \( f_x(x,y) \) at the point \( (-1, 4) \).
Step 4: Evaluate \( f_y(x,y) \) at the point \( (1, -4) \).
**Explanation of Procedures:**
1. **Partial Derivative with Respect to \( x \):**
To differentiate \( f(x,y) = \frac{5x^2 y}{e^{xy} + 3} \) with respect to \( x \), use the quotient rule:
\[
f_x(x,y) = \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}
\]
where \( u = 5x^2 y \) and \( v = e^{xy} + 3 \).
- First, find \( u' \) and \( v' \):
\[ u' = \frac{d}{dx} (5x^2 y) = 10xy \]
\[ v' = \frac{d}{dx} (e^{xy} + 3) = ye^{xy} \]
- Using the quotient rule:
\[
f_x(x,y) = \frac{(10xy)(e^{xy} + 3) - (5x^2 y)(ye^{xy})}{(e^{xy} + 3)^2}
\]
2. **Partial Derivative with Respect to \( y \):**
To differentiate \( f(x,y) \) with respect to \( y \):
-
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