Differentiate the function √√7e³x - 1 2x² + 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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Question
**Differentiate the Function**

Given function:

\[
\frac{\sqrt{7e^{3x} - 1}}{2x^2 + 1}
\]

To solve this problem, apply the quotient rule and chain rule of differentiation. Let \( u(x) = \sqrt{7e^{3x} - 1} \) and \( v(x) = 2x^2 + 1 \). Use the formula for the derivative of a quotient:

\[
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}
\]
Transcribed Image Text:**Differentiate the Function** Given function: \[ \frac{\sqrt{7e^{3x} - 1}}{2x^2 + 1} \] To solve this problem, apply the quotient rule and chain rule of differentiation. Let \( u(x) = \sqrt{7e^{3x} - 1} \) and \( v(x) = 2x^2 + 1 \). Use the formula for the derivative of a quotient: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]
Expert Solution
Step 1: Formula used

(u/v)' = (v.u'-uv')/v2 

(u+v)'= u'+v'  

(u.v)' = uv'+u'v  

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