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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Concept explainers
Rate of Change
The relation between two quantities which displays how much greater one quantity is than another is called ratio.
Slope
The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:
Question
![### Problem Statement
Find the gradient of the tangent line to the function \( y = \frac{3}{x^2} \) when \( x = 2 \).
---
### Solution Approach
1. **Understand the Function:**
- The function given is \( y = \frac{3}{x^2} \).
- This is a rational function where the variable \( x \) appears in the denominator squared.
2. **Find the Derivative:**
- The gradient of the tangent line to the curve at any point is given by the derivative of the function at that point.
- To find the derivative \( y' \) or \( \frac{dy}{dx} \), we use the differentiation rules.
3. **Differentiate the Function:**
- Using the power rule and chain rule, \( y = \frac{3}{x^2} \) can be rewritten as \( y = 3x^{-2} \).
- Differentiate \( y = 3x^{-2} \):
\[
y' = \frac{d}{dx}(3x^{-2}) = 3 \cdot (-2)x^{-3} = -6x^{-3}
\]
- Simplify the derivative:
\[
y' = \frac{-6}{x^3}
\]
4. **Calculate the Gradient at \( x = 2 \):**
- Substitute \( x = 2 \) into the derivative to find the gradient of the tangent line at this point:
\[
y' = \frac{-6}{(2)^3} = \frac{-6}{8} = -\frac{3}{4}
\]
### Final Answer
The gradient of the tangent line to the function \( y = \frac{3}{x^2} \) when \( x = 2 \) is \( -\frac{3}{4} \).
---
This solution illustrates the method of finding the tangent line's gradient to a given rational function at a specific point. The process involves differentiation and substitution, critical skills in calculus for analyzing the behavior of functions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec8465cd-fdc1-4278-9e3f-7f6c30db5e12%2F0633b40e-2cda-4c41-a749-a0cebb7816e1%2Fal7qg7zj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Find the gradient of the tangent line to the function \( y = \frac{3}{x^2} \) when \( x = 2 \).
---
### Solution Approach
1. **Understand the Function:**
- The function given is \( y = \frac{3}{x^2} \).
- This is a rational function where the variable \( x \) appears in the denominator squared.
2. **Find the Derivative:**
- The gradient of the tangent line to the curve at any point is given by the derivative of the function at that point.
- To find the derivative \( y' \) or \( \frac{dy}{dx} \), we use the differentiation rules.
3. **Differentiate the Function:**
- Using the power rule and chain rule, \( y = \frac{3}{x^2} \) can be rewritten as \( y = 3x^{-2} \).
- Differentiate \( y = 3x^{-2} \):
\[
y' = \frac{d}{dx}(3x^{-2}) = 3 \cdot (-2)x^{-3} = -6x^{-3}
\]
- Simplify the derivative:
\[
y' = \frac{-6}{x^3}
\]
4. **Calculate the Gradient at \( x = 2 \):**
- Substitute \( x = 2 \) into the derivative to find the gradient of the tangent line at this point:
\[
y' = \frac{-6}{(2)^3} = \frac{-6}{8} = -\frac{3}{4}
\]
### Final Answer
The gradient of the tangent line to the function \( y = \frac{3}{x^2} \) when \( x = 2 \) is \( -\frac{3}{4} \).
---
This solution illustrates the method of finding the tangent line's gradient to a given rational function at a specific point. The process involves differentiation and substitution, critical skills in calculus for analyzing the behavior of functions.
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