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
4.5 5
![**Problem Statement**
Find the derivative of the function.
\[ y = -6x \ln(5x + 8) \]
\[ \frac{dy}{dx} = \Box \]
**Explanation**
To find the derivative of the function \( y = -6x \ln(5x + 8) \), we will use the product rule and the chain rule of differentiation. The product rule states that for two functions \( u(x) \) and \( v(x) \), the derivative of their product is given by:
\[ (uv)' = u'v + uv' \]
Here, let \( u(x) = -6x \) and \( v(x) = \ln(5x + 8) \). Then, find the derivatives \( u'(x) \) and \( v'(x) \).
1. Derivative of \( u(x) \):
\[ u'(x) = -6 \]
2. Derivative of \( v(x) \) using the chain rule:
The derivative of \( \ln(5x + 8) \) is \( \frac{1}{5x + 8} \times 5 \) (since the derivative of the inside function, \( 5x + 8 \), is 5).
\[ v'(x) = \frac{5}{5x + 8} \]
Apply the product rule:
\[ \frac{dy}{dx} = (-6)(\ln(5x + 8)) + (-6x)\left(\frac{5}{5x + 8}\right) \]
Simplify to get the final derivative expression.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F34d89729-f0b7-4660-b47f-8f1bb9a32e44%2F2102f28f-775e-472e-a3b2-a4ad28f760c3%2Fder0oow_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
Find the derivative of the function.
\[ y = -6x \ln(5x + 8) \]
\[ \frac{dy}{dx} = \Box \]
**Explanation**
To find the derivative of the function \( y = -6x \ln(5x + 8) \), we will use the product rule and the chain rule of differentiation. The product rule states that for two functions \( u(x) \) and \( v(x) \), the derivative of their product is given by:
\[ (uv)' = u'v + uv' \]
Here, let \( u(x) = -6x \) and \( v(x) = \ln(5x + 8) \). Then, find the derivatives \( u'(x) \) and \( v'(x) \).
1. Derivative of \( u(x) \):
\[ u'(x) = -6 \]
2. Derivative of \( v(x) \) using the chain rule:
The derivative of \( \ln(5x + 8) \) is \( \frac{1}{5x + 8} \times 5 \) (since the derivative of the inside function, \( 5x + 8 \), is 5).
\[ v'(x) = \frac{5}{5x + 8} \]
Apply the product rule:
\[ \frac{dy}{dx} = (-6)(\ln(5x + 8)) + (-6x)\left(\frac{5}{5x + 8}\right) \]
Simplify to get the final derivative expression.
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