Find the derivative of the function. y= 2x°e* Which of the following shows how to find the derivative of the function dx dx dx O B. dx y' = 2x dx dx OC. y' = (2x°) ? O D. y'= 2x° (2*) . xp he derivative is y' =. %3D
Find the derivative of the function. y= 2x°e* Which of the following shows how to find the derivative of the function dx dx dx O B. dx y' = 2x dx dx OC. y' = (2x°) ? O D. y'= 2x° (2*) . xp he derivative is y' =. %3D
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
Need help
![**Find the Derivative of the Function**
Given: \( y = 2x^9 e^{x} \)
**Question:**
Which of the following shows how to find the derivative of the function?
---
**Options:**
- **A.**
\[
y' = 2x^9 \left(\frac{d}{dx} (2x^9)\right) + \left(\frac{d}{dx} (e^x)\right) e^x
\]
- **B.**
\[
y' = \frac{e^x \left(\frac{d}{dx} (2x^9)\right) - 2x^9 \left(\frac{d}{dx} (e^x)\right)}{(e^x)^2}
\]
- **C.**
\[
y' = \frac{2x^9 \left(\frac{d}{dx} (e^x)\right) - e^x \left(\frac{d}{dx} (2x^9)\right)}{(2x^9)^2}
\]
- **D.**
\[
y' = 2x^9 \left(\frac{d}{dx} (e^x)\right) + \left(\frac{d}{dx} (2x^9)\right) e^x
\]
---
**Correct Derivative:** \( y' = \boxed{\text{_____}} \)
**Explanation:**
This is a problem about finding the derivative of a function using basic differentiation rules. It involves a product of terms where you apply the product rule: \((uv)' = u'v + uv'\). You differentiate each part and then sum the results accordingly.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffc533774-2db2-4c40-ad62-e47c502b55d9%2F4751aa4a-8f26-4e90-a85c-a3ee3ba699cd%2Fvsf0vdk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Find the Derivative of the Function**
Given: \( y = 2x^9 e^{x} \)
**Question:**
Which of the following shows how to find the derivative of the function?
---
**Options:**
- **A.**
\[
y' = 2x^9 \left(\frac{d}{dx} (2x^9)\right) + \left(\frac{d}{dx} (e^x)\right) e^x
\]
- **B.**
\[
y' = \frac{e^x \left(\frac{d}{dx} (2x^9)\right) - 2x^9 \left(\frac{d}{dx} (e^x)\right)}{(e^x)^2}
\]
- **C.**
\[
y' = \frac{2x^9 \left(\frac{d}{dx} (e^x)\right) - e^x \left(\frac{d}{dx} (2x^9)\right)}{(2x^9)^2}
\]
- **D.**
\[
y' = 2x^9 \left(\frac{d}{dx} (e^x)\right) + \left(\frac{d}{dx} (2x^9)\right) e^x
\]
---
**Correct Derivative:** \( y' = \boxed{\text{_____}} \)
**Explanation:**
This is a problem about finding the derivative of a function using basic differentiation rules. It involves a product of terms where you apply the product rule: \((uv)' = u'v + uv'\). You differentiate each part and then sum the results accordingly.
Expert Solution
Step 1
Topic :- Derivative
Correct option :- D
Step by step
Solved in 2 steps
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning