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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![**Problem Statement:**
Prove, by showing all steps, that the derivative of \( \text{arccos} \, x \) is \( -\frac{1}{\sqrt{1-x^2}} \).
**Explanation:**
To prove this, we begin by using implicit differentiation. Let \( y = \text{arccos} \, x \). By definition of the arccosine function, we know that:
\[ \cos y = x \]
**Step-by-Step Derivation:**
1. Differentiate both sides of the equation \( \cos y = x \) with respect to \( x \).
2. The derivative of \( \cos y \) with respect to \( y \) is \( -\sin y \). By the chain rule, we have:
\[ \frac{d}{dx}(\cos y) = -\sin y \cdot \frac{dy}{dx} \]
3. The derivative of \( x \) with respect to \( x \) is 1, so:
\[ \frac{d}{dx}(x) = 1 \]
4. Set the derivatives equal to each other:
\[ -\sin y \cdot \frac{dy}{dx} = 1 \]
5. Solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{1}{\sin y} \]
6. Use the Pythagorean identity \( \sin^2 y + \cos^2 y = 1 \). Since \( \cos y = x \), then:
\[ \sin^2 y = 1 - \cos^2 y = 1 - x^2 \]
7. Therefore, \( \sin y = \sqrt{1 - x^2} \).
8. Substitute \( \sin y = \sqrt{1 - x^2} \) into the expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \]
Hence, the derivative of \( \text{arccos} \, x \) is \( -\frac{1}{\sqrt{1-x^2}} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F17a10252-ae51-4d74-b5c4-8b720ae84599%2Fe17be7c7-4e4d-4f10-bada-c7108a219704%2Flwbts3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Prove, by showing all steps, that the derivative of \( \text{arccos} \, x \) is \( -\frac{1}{\sqrt{1-x^2}} \).
**Explanation:**
To prove this, we begin by using implicit differentiation. Let \( y = \text{arccos} \, x \). By definition of the arccosine function, we know that:
\[ \cos y = x \]
**Step-by-Step Derivation:**
1. Differentiate both sides of the equation \( \cos y = x \) with respect to \( x \).
2. The derivative of \( \cos y \) with respect to \( y \) is \( -\sin y \). By the chain rule, we have:
\[ \frac{d}{dx}(\cos y) = -\sin y \cdot \frac{dy}{dx} \]
3. The derivative of \( x \) with respect to \( x \) is 1, so:
\[ \frac{d}{dx}(x) = 1 \]
4. Set the derivatives equal to each other:
\[ -\sin y \cdot \frac{dy}{dx} = 1 \]
5. Solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{1}{\sin y} \]
6. Use the Pythagorean identity \( \sin^2 y + \cos^2 y = 1 \). Since \( \cos y = x \), then:
\[ \sin^2 y = 1 - \cos^2 y = 1 - x^2 \]
7. Therefore, \( \sin y = \sqrt{1 - x^2} \).
8. Substitute \( \sin y = \sqrt{1 - x^2} \) into the expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \]
Hence, the derivative of \( \text{arccos} \, x \) is \( -\frac{1}{\sqrt{1-x^2}} \).
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