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
100%
![**Finding the Tangent Line to the Curve**
To find the equation of the tangent line to the curve at a given point, follow these steps:
Given:
\[ y = x^3 \ln(x) \]
Point: \((1, 0)\)
1. **Find the derivative of \(y = x^3 \ln(x)\):**
Using the product rule:
\[
y = u \cdot v, \text{ where } u = x^3 \text{ and } v = \ln(x)
\]
The product rule states:
\[
\frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v'
\]
Calculate \(u'\):
\[
u' = \frac{d}{dx}(x^3) = 3x^2
\]
Calculate \(v'\):
\[
v' = \frac{d}{dx}(\ln(x)) = \frac{1}{x}
\]
Substitute into the product rule:
\[
\frac{d}{dx}(x^3 \ln(x)) = 3x^2 \ln(x) + x^3 \cdot \frac{1}{x} = 3x^2 \ln(x) + x^2
\]
2. **Evaluate the derivative at the given point \( (1, 0) \):**
Plug \( x = 1 \) into the derivative:
\[
y' = 3(1)^2 \ln(1) + (1)^2 = 3 \cdot 0 + 1 = 1
\]
Therefore, the slope of the tangent line at \( (1, 0) \) is 1.
3. **Write the equation of the tangent line using the point-slope form of a line:**
The point-slope form is:
\[
y - y_1 = m(x - x_1)
\]
Substitute \( m = 1 \) and the point \( (1, 0) \):
\[
y - 0 = 1(x - 1)
\]
Simplify:
\[
y = x - 1
\]
Thus, the equation of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc1c90340-a748-4160-8298-b9e689632892%2Faf53eb34-2e27-4f47-a3ac-c0574a8f89f3%2F47j098b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Finding the Tangent Line to the Curve**
To find the equation of the tangent line to the curve at a given point, follow these steps:
Given:
\[ y = x^3 \ln(x) \]
Point: \((1, 0)\)
1. **Find the derivative of \(y = x^3 \ln(x)\):**
Using the product rule:
\[
y = u \cdot v, \text{ where } u = x^3 \text{ and } v = \ln(x)
\]
The product rule states:
\[
\frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v'
\]
Calculate \(u'\):
\[
u' = \frac{d}{dx}(x^3) = 3x^2
\]
Calculate \(v'\):
\[
v' = \frac{d}{dx}(\ln(x)) = \frac{1}{x}
\]
Substitute into the product rule:
\[
\frac{d}{dx}(x^3 \ln(x)) = 3x^2 \ln(x) + x^3 \cdot \frac{1}{x} = 3x^2 \ln(x) + x^2
\]
2. **Evaluate the derivative at the given point \( (1, 0) \):**
Plug \( x = 1 \) into the derivative:
\[
y' = 3(1)^2 \ln(1) + (1)^2 = 3 \cdot 0 + 1 = 1
\]
Therefore, the slope of the tangent line at \( (1, 0) \) is 1.
3. **Write the equation of the tangent line using the point-slope form of a line:**
The point-slope form is:
\[
y - y_1 = m(x - x_1)
\]
Substitute \( m = 1 \) and the point \( (1, 0) \):
\[
y - 0 = 1(x - 1)
\]
Simplify:
\[
y = x - 1
\]
Thus, the equation of
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images
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