Differentiate y = (2x + 1)°(x³ – x + 1)*. Solution: In this example we must use the Product Rule before using the Chain Rule: dy (2.x + 1) d (x³ – x + 1)* + (x³ – x + 1) · dx d (2x + 1) dx || dx d = (2x + 1) • 4(x³ – x + 1)³ (x³ – x + 1) dx

Differentiate y = (2x + 1)°(x³ – x + 1). Solution: In this example we must use the Product Rule before using the Chain Rule: dy (2.x + 1) d (x³ – x + 1) + (x³ – x + 1) · dx d (2x + 1) dx || dx d = (2x + 1) • 4(x³ – x + 1)³ (x³ – x + 1) dx

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

I don't understand how each step was obtained to get each value, could you please break down this example as to how each value was obtained because it's very confusing. I have attached the examples in screenshots.

Example 6
Differentiate y = (2x + 1)°(x³ – x + 1)*.
-
Solution:
In this example we must use the Product Rule before using
the Chain Rule:
d
dy
(2x + 1)'
(x³
– x + 1)* + (x³ – x + 1)+
d
(2x + 1)°
dx
|
-
dx
dx
d
(x³ – x + 1)
dx
(2x + 1) · 4(x³ – x + 1)³
.3
-
-
-
+ (x³ – x + 1)ª · 5(2x + 1)ª
d
(2x + 1)
dx
-

Example 6 – Solution
cont'd
= 4(2x + 1)°(x³ – x + 1)°(3x² – 1) + 5(x³ – x + 1)*(2x + 1)* · 2
-
Noticing that each term has the common factor
2(2x + 1)*(x³ – x + 1)³, we could factor it out and write the
-
answer as
dy
2(2.x + 1)*(x³ – x + 1)°(17x³ + 6x² – 9x + 3)
dx

Expert Solution

Step 1

Given:-

y =(2x+1)⁵(x³-x+1)⁴

To find:-

Derivative of y

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