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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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
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![### Evaluate the Integral
Evaluate the integral:
\[ \int \frac{dx}{x \sqrt{x^2 - 1}} \]
This problem involves finding the antiderivative of the given function. The integrand consists of a fraction where the numerator is \( dx \) and the denominator is \( x \sqrt{x^2 - 1} \). This can often be approached using trigonometric substitution or other integrative techniques based on the structure of the integrand.
#### Potential Solution Approach
1. **Identify the substitution**: Use a trigonometric substitution where \( x = \sec(\theta) \) which implies \( dx = \sec(\theta)\tan(\theta) d\theta \).
2. **Simplify the integrand** by substituting \( x \) and \( dx \) with the trigonometric identities.
3. **Compute the integral** in terms of the new variable \( \theta \).
4. **Reverse the substitution** to present the answer in terms of \( x \).
For more details on steps and techniques, refer to the section on trigonometric integrals and substitutions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1e2c0fb4-fe7c-46b9-9b1b-ce7222ffa53b%2Fc7b3e76c-6288-4e56-856c-4fe5eb9fb592%2Fhxt68eb8_processed.png&w=3840&q=75)
Transcribed Image Text:### Evaluate the Integral
Evaluate the integral:
\[ \int \frac{dx}{x \sqrt{x^2 - 1}} \]
This problem involves finding the antiderivative of the given function. The integrand consists of a fraction where the numerator is \( dx \) and the denominator is \( x \sqrt{x^2 - 1} \). This can often be approached using trigonometric substitution or other integrative techniques based on the structure of the integrand.
#### Potential Solution Approach
1. **Identify the substitution**: Use a trigonometric substitution where \( x = \sec(\theta) \) which implies \( dx = \sec(\theta)\tan(\theta) d\theta \).
2. **Simplify the integrand** by substituting \( x \) and \( dx \) with the trigonometric identities.
3. **Compute the integral** in terms of the new variable \( \theta \).
4. **Reverse the substitution** to present the answer in terms of \( x \).
For more details on steps and techniques, refer to the section on trigonometric integrals and substitutions.
![**Problem Statement:**
Find the derivative of \( y = \sin^{-1}(1 + x) \).
**Explanation:**
In this problem, we are tasked with finding the derivative of the inverse sine function, where the argument of the inverse sine function is \(1 + x\). The function is given as:
\[ y = \sin^{-1}(1 + x) \]
To find the derivative, we will need to use the chain rule. The chain rule is a formula for computing the derivative of the composition of two or more functions.
**Steps to Find the Derivative:**
1. **Identify the outer and inner functions:**
- The outer function is the inverse sine function: \( \sin^{-1}(u) \)
- The inner function is: \( u = 1 + x \)
2. **Differentiate the outer function with respect to the inner function:**
The derivative of the inverse sine function \(\sin^{-1}(u)\) with respect to \(u\) is:
\[
\frac{d}{du} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}}
\]
3. **Differentiate the inner function with respect to \(x\):**
The derivative of \( 1 + x \) with respect to \( x \) is:
\[
\frac{du}{dx} = 1
\]
4. **Apply the chain rule to find the derivative of \( y \):**
Using the chain rule:
\[
\frac{dy}{dx} = \frac{d}{du} \sin^{-1}(u) \cdot \frac{du}{dx}
\]
Substituting \( u = 1 + x \) and the derivatives:
\[
\frac{dy}{dx} = \frac{1}{\sqrt{1 - (1 + x)^2}} \cdot 1
\]
5. **Simplify the expression:**
The argument of the square root in the denominator is:
\[
1 - (1 + x)^2 = 1 - (1 + 2x + x^2) = 1 - 1 - 2x - x^2 = -2x - x^2
\]
Therefore:
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1e2c0fb4-fe7c-46b9-9b1b-ce7222ffa53b%2Fc7b3e76c-6288-4e56-856c-4fe5eb9fb592%2Fdjr62j4_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the derivative of \( y = \sin^{-1}(1 + x) \).
**Explanation:**
In this problem, we are tasked with finding the derivative of the inverse sine function, where the argument of the inverse sine function is \(1 + x\). The function is given as:
\[ y = \sin^{-1}(1 + x) \]
To find the derivative, we will need to use the chain rule. The chain rule is a formula for computing the derivative of the composition of two or more functions.
**Steps to Find the Derivative:**
1. **Identify the outer and inner functions:**
- The outer function is the inverse sine function: \( \sin^{-1}(u) \)
- The inner function is: \( u = 1 + x \)
2. **Differentiate the outer function with respect to the inner function:**
The derivative of the inverse sine function \(\sin^{-1}(u)\) with respect to \(u\) is:
\[
\frac{d}{du} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}}
\]
3. **Differentiate the inner function with respect to \(x\):**
The derivative of \( 1 + x \) with respect to \( x \) is:
\[
\frac{du}{dx} = 1
\]
4. **Apply the chain rule to find the derivative of \( y \):**
Using the chain rule:
\[
\frac{dy}{dx} = \frac{d}{du} \sin^{-1}(u) \cdot \frac{du}{dx}
\]
Substituting \( u = 1 + x \) and the derivatives:
\[
\frac{dy}{dx} = \frac{1}{\sqrt{1 - (1 + x)^2}} \cdot 1
\]
5. **Simplify the expression:**
The argument of the square root in the denominator is:
\[
1 - (1 + x)^2 = 1 - (1 + 2x + x^2) = 1 - 1 - 2x - x^2 = -2x - x^2
\]
Therefore:
\[
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