**Using the Chain Rule to Differentiate** ### Problem Statement: Use the Chain Rule to find \(\frac{\partial T}{\partial x}\) if \[ T = f(a, b, c) = a^3 + abc + c^2 \] where: - \( a = e^{2x} \cos y \) - \( b = e^{2x} \sin y \) - \( c = e^x \cos(2y) \) ### Explanation: To find the partial derivative of \( T \) with respect to \( x \) (\(\frac{\partial T}{\partial x}\)), apply the Chain Rule. This involves differentiating \( T \) with respect to each variable \( a \), \( b \), and \( c \), and then multiplying each by the derivative of that variable with respect to \( x \). ### Steps: 1. **Differentiate \( T \) with respect to \( a \), \( b \), and \( c \):** - \(\frac{\partial T}{\partial a} = 3a^2 + bc\) - \(\frac{\partial T}{\partial b} = ac\) - \(\frac{\partial T}{\partial c} = ab + 2c\) 2. **Differentiate each variable with respect to \( x \):** - \(\frac{da}{dx} = 2e^{2x} \cos y\) - \(\frac{db}{dx} = 2e^{2x} \sin y\) - \(\frac{dc}{dx} = e^x \cos(2y) + e^x (-2 \sin(2y))\) 3. **Apply the Chain Rule:** \[ \frac{\partial T}{\partial x} = \frac{\partial T}{\partial a} \cdot \frac{da}{dx} + \frac{\partial T}{\partial b} \cdot \frac{db}{dx} + \frac{\partial T}{\partial c} \cdot \frac{dc}{dx} \] Using these steps, one can compute the partial derivative \( \frac{\partial T}{\partial x} \).
**Using the Chain Rule to Differentiate** ### Problem Statement: Use the Chain Rule to find \(\frac{\partial T}{\partial x}\) if \[ T = f(a, b, c) = a^3 + abc + c^2 \] where: - \( a = e^{2x} \cos y \) - \( b = e^{2x} \sin y \) - \( c = e^x \cos(2y) \) ### Explanation: To find the partial derivative of \( T \) with respect to \( x \) (\(\frac{\partial T}{\partial x}\)), apply the Chain Rule. This involves differentiating \( T \) with respect to each variable \( a \), \( b \), and \( c \), and then multiplying each by the derivative of that variable with respect to \( x \). ### Steps: 1. **Differentiate \( T \) with respect to \( a \), \( b \), and \( c \):** - \(\frac{\partial T}{\partial a} = 3a^2 + bc\) - \(\frac{\partial T}{\partial b} = ac\) - \(\frac{\partial T}{\partial c} = ab + 2c\) 2. **Differentiate each variable with respect to \( x \):** - \(\frac{da}{dx} = 2e^{2x} \cos y\) - \(\frac{db}{dx} = 2e^{2x} \sin y\) - \(\frac{dc}{dx} = e^x \cos(2y) + e^x (-2 \sin(2y))\) 3. **Apply the Chain Rule:** \[ \frac{\partial T}{\partial x} = \frac{\partial T}{\partial a} \cdot \frac{da}{dx} + \frac{\partial T}{\partial b} \cdot \frac{db}{dx} + \frac{\partial T}{\partial c} \cdot \frac{dc}{dx} \] Using these steps, one can compute the partial derivative \( \frac{\partial T}{\partial x} \).
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
![**Using the Chain Rule to Differentiate**
### Problem Statement:
Use the Chain Rule to find \(\frac{\partial T}{\partial x}\) if
\[ T = f(a, b, c) = a^3 + abc + c^2 \]
where:
- \( a = e^{2x} \cos y \)
- \( b = e^{2x} \sin y \)
- \( c = e^x \cos(2y) \)
### Explanation:
To find the partial derivative of \( T \) with respect to \( x \) (\(\frac{\partial T}{\partial x}\)), apply the Chain Rule. This involves differentiating \( T \) with respect to each variable \( a \), \( b \), and \( c \), and then multiplying each by the derivative of that variable with respect to \( x \).
### Steps:
1. **Differentiate \( T \) with respect to \( a \), \( b \), and \( c \):**
- \(\frac{\partial T}{\partial a} = 3a^2 + bc\)
- \(\frac{\partial T}{\partial b} = ac\)
- \(\frac{\partial T}{\partial c} = ab + 2c\)
2. **Differentiate each variable with respect to \( x \):**
- \(\frac{da}{dx} = 2e^{2x} \cos y\)
- \(\frac{db}{dx} = 2e^{2x} \sin y\)
- \(\frac{dc}{dx} = e^x \cos(2y) + e^x (-2 \sin(2y))\)
3. **Apply the Chain Rule:**
\[
\frac{\partial T}{\partial x} = \frac{\partial T}{\partial a} \cdot \frac{da}{dx} + \frac{\partial T}{\partial b} \cdot \frac{db}{dx} + \frac{\partial T}{\partial c} \cdot \frac{dc}{dx}
\]
Using these steps, one can compute the partial derivative \( \frac{\partial T}{\partial x} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe620d41c-dd63-496b-b84c-cb17355be0e7%2Fbeb7e446-b357-46a9-b114-a3c3c74da65e%2Fj0pjnz.jpeg&w=3840&q=75)
Transcribed Image Text:**Using the Chain Rule to Differentiate**
### Problem Statement:
Use the Chain Rule to find \(\frac{\partial T}{\partial x}\) if
\[ T = f(a, b, c) = a^3 + abc + c^2 \]
where:
- \( a = e^{2x} \cos y \)
- \( b = e^{2x} \sin y \)
- \( c = e^x \cos(2y) \)
### Explanation:
To find the partial derivative of \( T \) with respect to \( x \) (\(\frac{\partial T}{\partial x}\)), apply the Chain Rule. This involves differentiating \( T \) with respect to each variable \( a \), \( b \), and \( c \), and then multiplying each by the derivative of that variable with respect to \( x \).
### Steps:
1. **Differentiate \( T \) with respect to \( a \), \( b \), and \( c \):**
- \(\frac{\partial T}{\partial a} = 3a^2 + bc\)
- \(\frac{\partial T}{\partial b} = ac\)
- \(\frac{\partial T}{\partial c} = ab + 2c\)
2. **Differentiate each variable with respect to \( x \):**
- \(\frac{da}{dx} = 2e^{2x} \cos y\)
- \(\frac{db}{dx} = 2e^{2x} \sin y\)
- \(\frac{dc}{dx} = e^x \cos(2y) + e^x (-2 \sin(2y))\)
3. **Apply the Chain Rule:**
\[
\frac{\partial T}{\partial x} = \frac{\partial T}{\partial a} \cdot \frac{da}{dx} + \frac{\partial T}{\partial b} \cdot \frac{db}{dx} + \frac{\partial T}{\partial c} \cdot \frac{dc}{dx}
\]
Using these steps, one can compute the partial derivative \( \frac{\partial T}{\partial x} \).
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