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:**
Evaluate
\[ \frac{d}{dx} \left( \frac{\sin^{-1} x}{\tan^{-1} x} \right). \]
Do not simplify at all.
---
In this calculus problem, you are required to differentiate the given expression with respect to \(x\) without simplifying the resulting expression. The expression involves the inverse trigonometric functions sine inverse (arcsine) and tangent inverse (arctangent).
**Step-by-Step Explanation:**
To evaluate this derivative, we need to apply the quotient rule, which is given by:
\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}, \]
where \( u = \sin^{-1} x \) and \( v = \tan^{-1} x \).
Here’s the step-by-step process:
1. **Identify \( u \) and \( v \):**
\[ u = \sin^{-1} x \]
\[ v = \tan^{-1} x \]
2. **Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):**
\[ \frac{du}{dx} = \frac{d}{dx} (\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \]
\[ \frac{dv}{dx} = \frac{d}{dx} (\tan^{-1} x) = \frac{1}{1 + x^2} \]
3. **Apply the quotient rule:**
\[ \frac{d}{dx} \left( \frac{\sin^{-1} x}{\tan^{-1} x} \right) = \frac{\left( \tan^{-1} x \right) \left( \frac{1}{\sqrt{1 - x^2}} \right) - \left( \sin^{-1} x \right) \left( \frac{1}{1 + x^2} \right)}{\left( \tan^{-1} x \right)^2} \]
Thus, the differentiated expression is:
\[ \frac{d}{dx} \left( \frac{\sin^{-1} x}{\tan](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc9a5f5a5-4eae-476c-bcab-4a001ac27552%2F3266e5fd-1595-4ed7-b718-ed73aa38bf05%2Fud8ml0c_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Evaluate
\[ \frac{d}{dx} \left( \frac{\sin^{-1} x}{\tan^{-1} x} \right). \]
Do not simplify at all.
---
In this calculus problem, you are required to differentiate the given expression with respect to \(x\) without simplifying the resulting expression. The expression involves the inverse trigonometric functions sine inverse (arcsine) and tangent inverse (arctangent).
**Step-by-Step Explanation:**
To evaluate this derivative, we need to apply the quotient rule, which is given by:
\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}, \]
where \( u = \sin^{-1} x \) and \( v = \tan^{-1} x \).
Here’s the step-by-step process:
1. **Identify \( u \) and \( v \):**
\[ u = \sin^{-1} x \]
\[ v = \tan^{-1} x \]
2. **Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):**
\[ \frac{du}{dx} = \frac{d}{dx} (\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \]
\[ \frac{dv}{dx} = \frac{d}{dx} (\tan^{-1} x) = \frac{1}{1 + x^2} \]
3. **Apply the quotient rule:**
\[ \frac{d}{dx} \left( \frac{\sin^{-1} x}{\tan^{-1} x} \right) = \frac{\left( \tan^{-1} x \right) \left( \frac{1}{\sqrt{1 - x^2}} \right) - \left( \sin^{-1} x \right) \left( \frac{1}{1 + x^2} \right)}{\left( \tan^{-1} x \right)^2} \]
Thus, the differentiated expression is:
\[ \frac{d}{dx} \left( \frac{\sin^{-1} x}{\tan
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