Find 4ery 4e zy( 4ry-1) Question Help: OVideo

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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In this exercise, we need to find the partial derivative of the expression \(4e^{4xy}\) with respect to \(x\).

### Problem Statement:
Find \(\frac{\partial}{\partial x} 4e^{4xy}\).

### Solution:

The solution provided in the box is:

\[
\frac{4e^{4xy}(4xy - 1)}{x^2}
\]

### Explanation:

To solve this problem, you would typically apply the chain rule for differentiation. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

The original function, \(4e^{4xy}\), is an exponential function where the exponent is a function of both \(x\) and \(y\). You need to differentiate it with respect to \(x\), treating \(y\) as a constant.

### Additional Information:

- **Question Help: Video**
  - It seems like there is an option to view a video for additional help on solving this type of problem. This could provide a step-by-step tutorial or explanation.
Transcribed Image Text:In this exercise, we need to find the partial derivative of the expression \(4e^{4xy}\) with respect to \(x\). ### Problem Statement: Find \(\frac{\partial}{\partial x} 4e^{4xy}\). ### Solution: The solution provided in the box is: \[ \frac{4e^{4xy}(4xy - 1)}{x^2} \] ### Explanation: To solve this problem, you would typically apply the chain rule for differentiation. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. The original function, \(4e^{4xy}\), is an exponential function where the exponent is a function of both \(x\) and \(y\). You need to differentiate it with respect to \(x\), treating \(y\) as a constant. ### Additional Information: - **Question Help: Video** - It seems like there is an option to view a video for additional help on solving this type of problem. This could provide a step-by-step tutorial or explanation.
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