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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![### Calculating the Gradient of a Function
**Problem Statement:**
Calculate the gradient of \( f(x,y) = \sin(x^2 - y) \).
**Solution:**
- The gradient of a function \( f(x, y) \) is denoted by \( \nabla f \).
- It is represented as a vector consisting of its partial derivatives with respect to \( x \) and \( y \).
\[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]
For \( f(x, y) = \sin(x^2 - y) \):
1. **Partial derivative with respect to \( x \):**
\[ \frac{\partial f}{\partial x} = \cos(x^2 - y) \cdot \frac{d}{dx}(x^2 - y) \]
\[ \frac{d}{dx}(x^2 - y) = 2x \]
So,
\[ \frac{\partial f}{\partial x} = \cos(x^2 - y) \cdot 2x \]
2. **Partial derivative with respect to \( y \):**
\[ \frac{\partial f}{\partial y} = \cos(x^2 - y) \cdot \frac{d}{dy}(x^2 - y) \]
\[ \frac{d}{dy}(x^2 - y) = -1 \]
So,
\[ \frac{\partial f}{\partial y} = \cos(x^2 - y) \cdot (-1) \]
\[ \frac{\partial f}{\partial y} = -\cos(x^2 - y) \]
Hence, the gradient \( \nabla f \) is:
\[ \nabla f = \left( 2x \cos(x^2 - y), -\cos(x^2 - y) \right) \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0761ca0b-8a30-47ed-b77e-00687bf0302e%2F1f22e2cb-a120-4f3c-b08f-907965da969c%2Fvbf0ppz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculating the Gradient of a Function
**Problem Statement:**
Calculate the gradient of \( f(x,y) = \sin(x^2 - y) \).
**Solution:**
- The gradient of a function \( f(x, y) \) is denoted by \( \nabla f \).
- It is represented as a vector consisting of its partial derivatives with respect to \( x \) and \( y \).
\[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]
For \( f(x, y) = \sin(x^2 - y) \):
1. **Partial derivative with respect to \( x \):**
\[ \frac{\partial f}{\partial x} = \cos(x^2 - y) \cdot \frac{d}{dx}(x^2 - y) \]
\[ \frac{d}{dx}(x^2 - y) = 2x \]
So,
\[ \frac{\partial f}{\partial x} = \cos(x^2 - y) \cdot 2x \]
2. **Partial derivative with respect to \( y \):**
\[ \frac{\partial f}{\partial y} = \cos(x^2 - y) \cdot \frac{d}{dy}(x^2 - y) \]
\[ \frac{d}{dy}(x^2 - y) = -1 \]
So,
\[ \frac{\partial f}{\partial y} = \cos(x^2 - y) \cdot (-1) \]
\[ \frac{\partial f}{\partial y} = -\cos(x^2 - y) \]
Hence, the gradient \( \nabla f \) is:
\[ \nabla f = \left( 2x \cos(x^2 - y), -\cos(x^2 - y) \right) \]
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