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...
Related questions
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:
\[
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa8713e9b-73ac-4e4f-a322-df330d573c40%2F6163e84c-8870-4d21-aa0f-57ca557dfabb%2F9itftx9_processed.jpeg&w=3840&q=75)
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
Step by step
Solved in 3 steps with 2 images
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