Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
**Title:** Derivative of \( y = \arccos(x) \) Using Implicit Differentiation and a Right Triangle **Objective:** Show that the derivative of \( y = \arccos(x) \) is \( \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \) using implicit differentiation and an appropriately labeled right triangle. ### Explanation: To find the derivative of \( y = \arccos(x) \) with respect to \( x \): 1. **Implicit Differentiation:** - Start with the given equation: \[ y = \arccos(x) \] - This implies: \[ \cos(y) = x \] - Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}[\cos(y)] = \frac{d}{dx}[x] \] - Use the chain rule on the left side and the derivative of the cosine function: \[ -\sin(y) \cdot \frac{dy}{dx} = 1 \] - Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = -\frac{1}{\sin(y)} \] 2. **Using a Right Triangle:** - Remember the trigonometric identity for arccosine: \[ \cos(y) = x \implies y = \arccos(x) \] - \( \cos(y) = x \) defines an adjacent side of a right triangle with hypotenuse 1. - Using the Pythagorean theorem, the opposite side is given by: \[ \sin(y) = \sqrt{1 - x^2} \] - Substitute this back into the derivative expression: \[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \] Thus, the derivative of \( y = \arccos(x) \) is: \[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \] ### Graphical Representation: - **Right Triangle Diagram:** * **Hypotenuse
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Transcribed Image Text:**Title:** Derivative of \( y = \arccos(x) \) Using Implicit Differentiation and a Right Triangle **Objective:** Show that the derivative of \( y = \arccos(x) \) is \( \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \) using implicit differentiation and an appropriately labeled right triangle. ### Explanation: To find the derivative of \( y = \arccos(x) \) with respect to \( x \): 1. **Implicit Differentiation:** - Start with the given equation: \[ y = \arccos(x) \] - This implies: \[ \cos(y) = x \] - Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}[\cos(y)] = \frac{d}{dx}[x] \] - Use the chain rule on the left side and the derivative of the cosine function: \[ -\sin(y) \cdot \frac{dy}{dx} = 1 \] - Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = -\frac{1}{\sin(y)} \] 2. **Using a Right Triangle:** - Remember the trigonometric identity for arccosine: \[ \cos(y) = x \implies y = \arccos(x) \] - \( \cos(y) = x \) defines an adjacent side of a right triangle with hypotenuse 1. - Using the Pythagorean theorem, the opposite side is given by: \[ \sin(y) = \sqrt{1 - x^2} \] - Substitute this back into the derivative expression: \[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \] Thus, the derivative of \( y = \arccos(x) \) is: \[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \] ### Graphical Representation: - **Right Triangle Diagram:** * **Hypotenuse
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