Calculate fz (0, 0, 1, 1), where f(x, y, z, w) = exz+y z² + w
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem Statement
Calculate \( f_z(0, 0, 1, 1) \), where \( f(x, y, z, w) = \frac{e^{xz+y}}{z^2 + w} \).
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**Explanation:**
The problem requires calculating the partial derivative of the function \( f \) with respect to the variable \( z \) at the point \( (0, 0, 1, 1) \). The function \( f(x, y, z, w) \) is given as:
\[ f(x, y, z, w) = \frac{e^{xz + y}}{z^2 + w} \]
To solve this, you will need to follow these steps:
1. **Differentiate \( f \) with respect to \( z \).**
2. **Substitute \( x = 0 \), \( y = 0 \), \( z = 1 \), and \( w = 1 \) into the derivative.**
#### Differentiation Step-by-Step:
1. **Differentiate the numerator and the denominator separately and then apply the quotient rule:**
\[ \frac{\partial}{\partial z}\left(\frac{e^{xz + y}}{z^2 + w}\right) = \frac{ (z^2+w) \frac{\partial e^{xz+y}}{\partial z} - e^{xz+y} \frac{\partial (z^2+w)}{\partial z} }{(z^2+w)^2} \]
2. **Compute \(\frac{\partial e^{xz+y}}{\partial z}\):**
\[ \frac{\partial e^{xz + y}}{\partial z} = xe^{xz + y} \]
3. **Compute \(\frac{\partial (z^2 + w)}{\partial z}\):**
\[ \frac{\partial (z^2 + w)}{\partial z} = 2z \]
4. **Substitute these into the quotient rule expression:**
\[ \frac{\partial}{\partial z}\left(\frac{e^{xz + y}}{z^2 + w}\right) = \frac{(z^2 + w) \cdot (xe^{xz + y}) - e^{xz + y} \cdot 2z}{(z^2 + w)^2} \]
5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F717f4d87-5a32-4cdf-855f-90bb556622e3%2F3397412d-5f1f-448e-ba93-cae6a5e819aa%2Fon5v0m_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Calculate \( f_z(0, 0, 1, 1) \), where \( f(x, y, z, w) = \frac{e^{xz+y}}{z^2 + w} \).
---
**Explanation:**
The problem requires calculating the partial derivative of the function \( f \) with respect to the variable \( z \) at the point \( (0, 0, 1, 1) \). The function \( f(x, y, z, w) \) is given as:
\[ f(x, y, z, w) = \frac{e^{xz + y}}{z^2 + w} \]
To solve this, you will need to follow these steps:
1. **Differentiate \( f \) with respect to \( z \).**
2. **Substitute \( x = 0 \), \( y = 0 \), \( z = 1 \), and \( w = 1 \) into the derivative.**
#### Differentiation Step-by-Step:
1. **Differentiate the numerator and the denominator separately and then apply the quotient rule:**
\[ \frac{\partial}{\partial z}\left(\frac{e^{xz + y}}{z^2 + w}\right) = \frac{ (z^2+w) \frac{\partial e^{xz+y}}{\partial z} - e^{xz+y} \frac{\partial (z^2+w)}{\partial z} }{(z^2+w)^2} \]
2. **Compute \(\frac{\partial e^{xz+y}}{\partial z}\):**
\[ \frac{\partial e^{xz + y}}{\partial z} = xe^{xz + y} \]
3. **Compute \(\frac{\partial (z^2 + w)}{\partial z}\):**
\[ \frac{\partial (z^2 + w)}{\partial z} = 2z \]
4. **Substitute these into the quotient rule expression:**
\[ \frac{\partial}{\partial z}\left(\frac{e^{xz + y}}{z^2 + w}\right) = \frac{(z^2 + w) \cdot (xe^{xz + y}) - e^{xz + y} \cdot 2z}{(z^2 + w)^2} \]
5
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