18 Oct 2021
Problem 25
Page 143
Section 2.6: Derivatives and Rates of Change
Chapter 2: Limits and Derivatives
Textbook ExpertVerified Tutor
18 Oct 2021
Given information
The equation of the curve is 
and the given point is 
.
Step-by-step explanation
Step 1.
First find 
Therefore the slope of the tangent will be 
Now the equation of the tangent to the given curve and passing through the point is:
Single Variable Calculus: Early Transcendentals
4th Edition, 2018
Stewart
ISBN: 9781337687805
Solutions
Chapter
Section
- Problem 1a
- Problem 1b
- Problem 2
- Problem 3a
- Problem 3b
- Problem 3c
- Problem 4a
- Problem 4a
- Problem 4b
- Problem 4c
- Problem 5
- Problem 6
- Problem 7
- Problem 8
- Problem 9a
- Problem 9b
- Problem 9c
- Problem 10a
- Problem 10b
- Problem 10c
- Problem 11a
- Problem 11b
- Problem 12a
- Problem 12b
- Problem 12c
- Problem 13
- Problem 14a
- Problem 14b
- Problem 14c
- Problem 14d
- Problem 15
- Problem 16a
- Problem 16b
- Problem 16c
- Problem 17
- Problem 18
- Problem 19
- Problem 20
- Problem 21
- Problem 22
- Problem 23
- Problem 24
- Problem 25
- Problem 25a
- Problem 25b
- Problem 26a
- Problem 26b
- Problem 27
- Problem 28
- Problem 29
- Problem 30
- Problem 31
- Problem 32
- Problem 33
- Problem 34
- Problem 35
- Problem 36
- Problem 37
- Problem 38
- Problem 39
- Problem 40
- Problem 41
- Problem 42
- Problem 43a
- Problem 43b
- Problem 43c
- Problem 44
- Problem 44
- Problem 44
- Problem 44a
- Problem 44b
- Problem 44c
- Problem 45a
- Problem 45b
- Problem 46
- Problem 47a
- Problem 47b
- Problem 47c
- Problem 48a
- Problem 48b
- Problem 49
- Problem 50a
- Problem 50b
- Problem 51a
- Problem 51b
- Problem 52a
- Problem 52b
- Problem 53
- Problem 54