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
![To find the derivative \(\frac{dy}{dx}\) of the function \(y = \frac{x^3 + 4x + 7}{x^2 + 1}\), we will use the quotient rule. The quotient rule states that if you have a function in the form \(\frac{u}{v}\), then its derivative is given by:
\[
\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}
\]
For this specific function:
- Let \(u = x^3 + 4x + 7\)
- Let \(v = x^2 + 1\)
First, find the derivatives \(\frac{du}{dx}\) and \(\frac{dv}{dx}\):
\[
\frac{du}{dx} = 3x^2 + 4
\]
\[
\frac{dv}{dx} = 2x
\]
Now apply the quotient rule:
\[
\frac{dy}{dx} = \frac{(x^2 + 1)(3x^2 + 4) - (x^3 + 4x + 7)(2x)}{(x^2 + 1)^2}
\]
Simplify the expression to find the final derivative:
\[
\frac{dy}{dx} = \frac{(3x^4 + 4x^2 + 3x^2 + 4) - (2x^4 + 8x^2 + 14x)}{(x^2 + 1)^2}
\]
\[
= \frac{3x^4 + 7x^2 + 4 - 2x^4 - 8x^2 - 14x}{(x^2 + 1)^2}
\]
\[
= \frac{x^4 - x^2 - 14x + 4}{(x^2 + 1)^2}
\]
Thus, the derivative is:
\[
\frac{dy}{dx} = \frac{x^4 - x^2 - 14x + 4}{(x^2 + 1)^2}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3e51674d-3d29-4cb4-94ad-529601d10286%2Fcffce628-7eb1-4d7d-bdae-86440f7a846d%2F2ueqbvx_processed.png&w=3840&q=75)
Transcribed Image Text:To find the derivative \(\frac{dy}{dx}\) of the function \(y = \frac{x^3 + 4x + 7}{x^2 + 1}\), we will use the quotient rule. The quotient rule states that if you have a function in the form \(\frac{u}{v}\), then its derivative is given by:
\[
\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}
\]
For this specific function:
- Let \(u = x^3 + 4x + 7\)
- Let \(v = x^2 + 1\)
First, find the derivatives \(\frac{du}{dx}\) and \(\frac{dv}{dx}\):
\[
\frac{du}{dx} = 3x^2 + 4
\]
\[
\frac{dv}{dx} = 2x
\]
Now apply the quotient rule:
\[
\frac{dy}{dx} = \frac{(x^2 + 1)(3x^2 + 4) - (x^3 + 4x + 7)(2x)}{(x^2 + 1)^2}
\]
Simplify the expression to find the final derivative:
\[
\frac{dy}{dx} = \frac{(3x^4 + 4x^2 + 3x^2 + 4) - (2x^4 + 8x^2 + 14x)}{(x^2 + 1)^2}
\]
\[
= \frac{3x^4 + 7x^2 + 4 - 2x^4 - 8x^2 - 14x}{(x^2 + 1)^2}
\]
\[
= \frac{x^4 - x^2 - 14x + 4}{(x^2 + 1)^2}
\]
Thus, the derivative is:
\[
\frac{dy}{dx} = \frac{x^4 - x^2 - 14x + 4}{(x^2 + 1)^2}
\]
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