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
![**Problem Statement:**
Find \(\frac{dy}{dx}\) where \(y = \csc^{-1} \left (\frac{1}{6x + 9} \right )\).
**Instructions:**
Provide your answer below:
\(\frac{dy}{dx}\) = [ ]
---
**Explanation:**
Here, you are required to find the derivative of the function \(y\) with respect to \(x\). The function \(y\) is given as the inverse cosecant \(\left(\csc^{-1}\right)\) of the expression \(\frac{1}{6x + 9}\).
**Steps to Solution:**
1. **Rewrite the function**:
\[
y = \csc^{-1} \left (\frac{1}{6x + 9} \right )
\]
2. **Recall the derivative of inverse cosecant function**:
\[
\frac{d}{dx} \left( \csc^{-1}(u) \right) = \frac{-1}{|u| \sqrt{u^2 - 1}} \cdot \frac{du}{dx}
\]
3. **Substitute \(u = \frac{1}{6x + 9}\) and find \(\frac{du}{dx}\)**:
\[
u = \frac{1}{6x + 9}, \quad \frac{du}{dx} = \frac{d}{dx} \left( \frac{1}{6x + 9} \right) = \frac{-6}{(6x + 9)^2}
\]
4. **Apply the chain rule for differentiation**:
\[
\frac{dy}{dx} = \frac{-1}{\left| \frac{1}{6x + 9} \right| \sqrt{\left( \frac{1}{6x + 9} \right)^2 - 1}} \cdot \frac{-6}{(6x + 9)^2}
\]
5. **Simplify the expression** to obtain the final derivative.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5ea397a-632d-4020-8701-e2de67ac8a17%2F9416f593-38ea-41b6-acb6-0c630ff73ef0%2Fk9mgwrm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find \(\frac{dy}{dx}\) where \(y = \csc^{-1} \left (\frac{1}{6x + 9} \right )\).
**Instructions:**
Provide your answer below:
\(\frac{dy}{dx}\) = [ ]
---
**Explanation:**
Here, you are required to find the derivative of the function \(y\) with respect to \(x\). The function \(y\) is given as the inverse cosecant \(\left(\csc^{-1}\right)\) of the expression \(\frac{1}{6x + 9}\).
**Steps to Solution:**
1. **Rewrite the function**:
\[
y = \csc^{-1} \left (\frac{1}{6x + 9} \right )
\]
2. **Recall the derivative of inverse cosecant function**:
\[
\frac{d}{dx} \left( \csc^{-1}(u) \right) = \frac{-1}{|u| \sqrt{u^2 - 1}} \cdot \frac{du}{dx}
\]
3. **Substitute \(u = \frac{1}{6x + 9}\) and find \(\frac{du}{dx}\)**:
\[
u = \frac{1}{6x + 9}, \quad \frac{du}{dx} = \frac{d}{dx} \left( \frac{1}{6x + 9} \right) = \frac{-6}{(6x + 9)^2}
\]
4. **Apply the chain rule for differentiation**:
\[
\frac{dy}{dx} = \frac{-1}{\left| \frac{1}{6x + 9} \right| \sqrt{\left( \frac{1}{6x + 9} \right)^2 - 1}} \cdot \frac{-6}{(6x + 9)^2}
\]
5. **Simplify the expression** to obtain the final derivative.
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