Find dy dx dy dx if y = = x³ + 4x + 7 x² + 1

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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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:

\[
\frac{dy}{dx} = \frac{x^4 - x^2 - 14x + 4}{(x^2 + 1)^2}
\]
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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