Compute the first-order partial derivatives of the function. z = sinh (6x³y) (Use symbolic notation and fractions where needed.)
Compute the first-order partial derivatives of the function. z = sinh (6x³y) (Use symbolic notation and fractions where needed.)
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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![## Computing First-Order Partial Derivatives
### Problem Statement
Compute the first-order partial derivatives of the function:
\[ z = \sinh(6x^3 y) \]
(Use symbolic notation and fractions where needed.)
### Solution Attempts
1. **First Partial Derivative with Respect to \( x \)**:
\[
\frac{\partial z}{\partial x} = 18x^2 y \cos (6x^3 y)
\]
However, this response is marked as **Incorrect**.
2. **First Partial Derivative with Respect to \( y \)**:
\[
\frac{\partial z}{\partial y} = 6x^3 \cos (6x^3 y)
\]
This response is also marked as **Incorrect**.
### Notes:
- The notation \( \sinh(u) \) represents the hyperbolic sine function.
- It is important to consider the chain rule when differentiating functions of the form \( \sinh(g(x,y)) \).
### Graphs and Diagrams
There are no graphs or diagrams included in this problem statement.
### Correct Approach:
To correctly compute the partial derivatives, apply the chain rule appropriately.
#### Step-by-Step Solution:
1. **Compute \( \frac{\partial z}{\partial x} \)**:
\[
z = \sinh(6x^3 y)
\]
Let \( u = 6x^3 y \), then \( z = \sinh(u) \).
Using the chain rule:
\[
\frac{\partial z}{\partial x} = \frac{d \sinh(u)}{du} \cdot \frac{\partial u}{\partial x}
\]
We know:
\[
\frac{d \sinh(u)}{du} = \cosh(u)
\]
and
\[
\frac{\partial u}{\partial x} = \frac{\partial (6x^3 y)}{\partial x} = 18x^2 y
\]
Therefore:
\[
\frac{\partial z}{\partial x} = \cosh(6x^3 y) \cdot 18x^2 y
\]
2. **Compute \( \frac{\partial z}{\partial y} \)**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f28cfcf-b497-42da-bd62-a5ef374f1d5f%2F9e9dd129-21f5-4b98-8894-009ff2079631%2F5lzum7i_processed.png&w=3840&q=75)
Transcribed Image Text:## Computing First-Order Partial Derivatives
### Problem Statement
Compute the first-order partial derivatives of the function:
\[ z = \sinh(6x^3 y) \]
(Use symbolic notation and fractions where needed.)
### Solution Attempts
1. **First Partial Derivative with Respect to \( x \)**:
\[
\frac{\partial z}{\partial x} = 18x^2 y \cos (6x^3 y)
\]
However, this response is marked as **Incorrect**.
2. **First Partial Derivative with Respect to \( y \)**:
\[
\frac{\partial z}{\partial y} = 6x^3 \cos (6x^3 y)
\]
This response is also marked as **Incorrect**.
### Notes:
- The notation \( \sinh(u) \) represents the hyperbolic sine function.
- It is important to consider the chain rule when differentiating functions of the form \( \sinh(g(x,y)) \).
### Graphs and Diagrams
There are no graphs or diagrams included in this problem statement.
### Correct Approach:
To correctly compute the partial derivatives, apply the chain rule appropriately.
#### Step-by-Step Solution:
1. **Compute \( \frac{\partial z}{\partial x} \)**:
\[
z = \sinh(6x^3 y)
\]
Let \( u = 6x^3 y \), then \( z = \sinh(u) \).
Using the chain rule:
\[
\frac{\partial z}{\partial x} = \frac{d \sinh(u)}{du} \cdot \frac{\partial u}{\partial x}
\]
We know:
\[
\frac{d \sinh(u)}{du} = \cosh(u)
\]
and
\[
\frac{\partial u}{\partial x} = \frac{\partial (6x^3 y)}{\partial x} = 18x^2 y
\]
Therefore:
\[
\frac{\partial z}{\partial x} = \cosh(6x^3 y) \cdot 18x^2 y
\]
2. **Compute \( \frac{\partial z}{\partial y} \)**
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