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: Differentiation**
Differentiate the function \( f(x) = \sin^{-1}(4x) \). Use exact values.
\[ f'(x) = \]
(Note: The image shows a problem in calculus, specifically focused on finding the derivative of an inverse sine function. There are no graphs or additional diagrams in the image.)
**Solution:**
To differentiate \( f(x) = \sin^{-1}(4x) \), we use the chain rule. The general formula for the derivative of \( \sin^{-1}(u) \) where \( u \) is a function of \( x \) is:
\[ \frac{d}{dx} [\sin^{-1}(u)] = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
In this case, \( u = 4x \).
So, we find \( \frac{du}{dx} \):
\[ \frac{d}{dx} [4x] = 4 \]
Now, apply the formula:
\[ f'(x) = \frac{1}{\sqrt{1 - (4x)^2}} \cdot 4 \]
Simplify the expression:
\[ f'(x) = \frac{4}{\sqrt{1 - 16x^2}} \]
Thus, the derivative of \( f(x) = \sin^{-1}(4x) \) is:
\[ f'(x) = \frac{4}{\sqrt{1 - 16x^2}} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19a44d99-9126-45f2-b575-2ff09f428fba%2F6c62835c-5125-40ba-b618-85fe55fd9195%2Fi17zurr.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement: Differentiation**
Differentiate the function \( f(x) = \sin^{-1}(4x) \). Use exact values.
\[ f'(x) = \]
(Note: The image shows a problem in calculus, specifically focused on finding the derivative of an inverse sine function. There are no graphs or additional diagrams in the image.)
**Solution:**
To differentiate \( f(x) = \sin^{-1}(4x) \), we use the chain rule. The general formula for the derivative of \( \sin^{-1}(u) \) where \( u \) is a function of \( x \) is:
\[ \frac{d}{dx} [\sin^{-1}(u)] = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
In this case, \( u = 4x \).
So, we find \( \frac{du}{dx} \):
\[ \frac{d}{dx} [4x] = 4 \]
Now, apply the formula:
\[ f'(x) = \frac{1}{\sqrt{1 - (4x)^2}} \cdot 4 \]
Simplify the expression:
\[ f'(x) = \frac{4}{\sqrt{1 - 16x^2}} \]
Thus, the derivative of \( f(x) = \sin^{-1}(4x) \) is:
\[ f'(x) = \frac{4}{\sqrt{1 - 16x^2}} \]
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