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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how do i solve the attached question about the chain rule.
![**Calculate the Derivative Using Implicit Differentiation:**
Given the equation:
\[ x^3 w + w^8 + w z^2 + 7yz = 0 \]
Find the derivative \(\frac{\partial w}{\partial z}\).
**Solution:**
To solve this problem, we need to implicitly differentiate each term of the equation with respect to \(z\) while treating \(w\) as a function of \(z\).
1. Differentiate \(x^3 w\) with respect to \(z\):
- Since \(x^3\) is a constant with respect to \(z\), use the product rule: \(\frac{\partial}{\partial z}(x^3 w) = x^3 \frac{\partial w}{\partial z}\).
2. Differentiate \(w^8\) with respect to \(z\):
- Use the chain rule: \(\frac{\partial}{\partial z}(w^8) = 8w^7 \frac{\partial w}{\partial z}\).
3. Differentiate \(w z^2\) with respect to \(z\):
- Also use the product rule: \(\frac{\partial}{\partial z}(w z^2) = z^2 \frac{\partial w}{\partial z} + w \cdot 2z\).
4. Differentiate \(7yz\) with respect to \(z\):
- Notice that \(y\) is treated as a constant: \(\frac{\partial}{\partial z}(7yz) = 7y\).
Substitute these derivatives back into the differentiated equation:
\[ x^3 \frac{\partial w}{\partial z} + 8w^7 \frac{\partial w}{\partial z} + z^2 \frac{\partial w}{\partial z} + 2wz + 7y = 0 \]
Now, solve for \(\frac{\partial w}{\partial z}\) by combining like terms:
\[ (x^3 + 8w^7 + z^2) \frac{\partial w}{\partial z} = -2wz - 7y \]
Thus, the explicit form of \(\frac{\partial w}{\partial z}\) is:
\[ \frac{\partial w}{\partial z} = \frac{-2wz](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb3cfeff4-8ba6-46a8-98d4-804b4f4f620a%2Fad5885aa-6b79-489d-9ad5-23d68c72a0b3%2Frorlekh_processed.png&w=3840&q=75)
Transcribed Image Text:**Calculate the Derivative Using Implicit Differentiation:**
Given the equation:
\[ x^3 w + w^8 + w z^2 + 7yz = 0 \]
Find the derivative \(\frac{\partial w}{\partial z}\).
**Solution:**
To solve this problem, we need to implicitly differentiate each term of the equation with respect to \(z\) while treating \(w\) as a function of \(z\).
1. Differentiate \(x^3 w\) with respect to \(z\):
- Since \(x^3\) is a constant with respect to \(z\), use the product rule: \(\frac{\partial}{\partial z}(x^3 w) = x^3 \frac{\partial w}{\partial z}\).
2. Differentiate \(w^8\) with respect to \(z\):
- Use the chain rule: \(\frac{\partial}{\partial z}(w^8) = 8w^7 \frac{\partial w}{\partial z}\).
3. Differentiate \(w z^2\) with respect to \(z\):
- Also use the product rule: \(\frac{\partial}{\partial z}(w z^2) = z^2 \frac{\partial w}{\partial z} + w \cdot 2z\).
4. Differentiate \(7yz\) with respect to \(z\):
- Notice that \(y\) is treated as a constant: \(\frac{\partial}{\partial z}(7yz) = 7y\).
Substitute these derivatives back into the differentiated equation:
\[ x^3 \frac{\partial w}{\partial z} + 8w^7 \frac{\partial w}{\partial z} + z^2 \frac{\partial w}{\partial z} + 2wz + 7y = 0 \]
Now, solve for \(\frac{\partial w}{\partial z}\) by combining like terms:
\[ (x^3 + 8w^7 + z^2) \frac{\partial w}{\partial z} = -2wz - 7y \]
Thus, the explicit form of \(\frac{\partial w}{\partial z}\) is:
\[ \frac{\partial w}{\partial z} = \frac{-2wz
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