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Compute the derivatives of the given function. d [cos ¹(√)] dx - D

Compute the derivatives of the given function. d [cos ¹(√)] dx - D

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.CR: Chapter 4 Review
Problem 4CR: Determine whether each of the following statements is true or false, and explain why. The derivative...
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**Problem Statement:** Compute the derivatives of the given function. \[ \frac{d}{dx} \left[ \cos^{-1} (\sqrt{x}) \right] = \] **Explanation:** - The problem requires you to compute the derivative of the inverse cosine function which has an argument of the square root of \( x \). - The inverse cosine function is denoted by \( \cos^{-1} \). To solve this problem, use the chain rule for differentiation. The chain rule states: \[ \frac{d}{dx} \left[ f(g(x)) \right] = f'(g(x)) \cdot g'(x) \] In this case, let: \[ f(u) = \cos^{-1} (u) \] \[ g(x) = \sqrt{x} \] First, differentiate \( f(u) \): \[ \frac{d}{du} \left[ \cos^{-1}(u) \right] = -\frac{1}{\sqrt{1 - u^2}} \] Next, differentiate \( g(x) \): \[ \frac{d}{dx} [\sqrt{x}] = \frac{1}{2\sqrt{x}} \] Now, apply the chain rule: \[ \frac{d}{dx} \left[ \cos^{-1} (\sqrt{x}) \right] = \left( -\frac{1}{\sqrt{1 - (\sqrt{x})^2}} \right) \left( \frac{1}{2 \sqrt{x}} \right) \] \[ = -\frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2 \sqrt{x}} \] \[ = -\frac{1}{2 \sqrt{x} \sqrt{1 - x}} \] \[ = -\frac{1}{2 \sqrt{x(1 - x)}} \] The derivative of the given function is: \[ \frac{d}{dx} \left[ \cos^{-1} (\sqrt{x}) \right] = -\frac{1}{2 \sqrt{x(1 - x)}} \]
Transcribed Image Text:**Problem Statement:** Compute the derivatives of the given function. \[ \frac{d}{dx} \left[ \cos^{-1} (\sqrt{x}) \right] = \] **Explanation:** - The problem requires you to compute the derivative of the inverse cosine function which has an argument of the square root of \( x \). - The inverse cosine function is denoted by \( \cos^{-1} \). To solve this problem, use the chain rule for differentiation. The chain rule states: \[ \frac{d}{dx} \left[ f(g(x)) \right] = f'(g(x)) \cdot g'(x) \] In this case, let: \[ f(u) = \cos^{-1} (u) \] \[ g(x) = \sqrt{x} \] First, differentiate \( f(u) \): \[ \frac{d}{du} \left[ \cos^{-1}(u) \right] = -\frac{1}{\sqrt{1 - u^2}} \] Next, differentiate \( g(x) \): \[ \frac{d}{dx} [\sqrt{x}] = \frac{1}{2\sqrt{x}} \] Now, apply the chain rule: \[ \frac{d}{dx} \left[ \cos^{-1} (\sqrt{x}) \right] = \left( -\frac{1}{\sqrt{1 - (\sqrt{x})^2}} \right) \left( \frac{1}{2 \sqrt{x}} \right) \] \[ = -\frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2 \sqrt{x}} \] \[ = -\frac{1}{2 \sqrt{x} \sqrt{1 - x}} \] \[ = -\frac{1}{2 \sqrt{x(1 - x)}} \] The derivative of the given function is: \[ \frac{d}{dx} \left[ \cos^{-1} (\sqrt{x}) \right] = -\frac{1}{2 \sqrt{x(1 - x)}} \]
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