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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4.1 #9
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![**Calculus Problem: Derivative of a Function**
**Problem Statement:**
Find the derivative of the function.
\[ y = \frac{4x^3 + 6}{x} \]
**Solution:**
To find the derivative \( y' \), we can first simplify the function by dividing each term in the numerator by the denominator:
\[ y = \frac{4x^3}{x} + \frac{6}{x} \]
Simplifying, we get:
\[ y = 4x^2 + 6x^{-1} \]
Now, we differentiate each term separately:
1. The derivative of \( 4x^2 \) is \( 8x \).
2. The derivative of \( 6x^{-1} \) is \(-6x^{-2}\).
Thus:
\[ y' = 8x - \frac{6}{x^2} \]
**Final Answer:**
\[ y' = 8x - \frac{6}{x^2} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63591bc8-75d6-42dc-963a-0ef4cf7b290a%2Fb35eb596-f559-436e-83c0-e2034bc57403%2Fvjnhph_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculus Problem: Derivative of a Function**
**Problem Statement:**
Find the derivative of the function.
\[ y = \frac{4x^3 + 6}{x} \]
**Solution:**
To find the derivative \( y' \), we can first simplify the function by dividing each term in the numerator by the denominator:
\[ y = \frac{4x^3}{x} + \frac{6}{x} \]
Simplifying, we get:
\[ y = 4x^2 + 6x^{-1} \]
Now, we differentiate each term separately:
1. The derivative of \( 4x^2 \) is \( 8x \).
2. The derivative of \( 6x^{-1} \) is \(-6x^{-2}\).
Thus:
\[ y' = 8x - \frac{6}{x^2} \]
**Final Answer:**
\[ y' = 8x - \frac{6}{x^2} \]
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