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
![**Mathematical Problem: Finding the Derivative**
Given the function:
\[ f(x) = -2 \sin^{-1} (e^{8x}) \]
We need to find the derivative, \( f'(x) \).
**Solution:**
To find \( f'(x) \), we apply the chain rule. The derivative of \(\sin^{-1}(u)\) with respect to \(u\) is \(\frac{1}{\sqrt{1-u^2}}\).
Given \( u = e^{8x} \), the derivative of \( e^{8x} \) with respect to \(x\) is \( 8e^{8x} \).
Thus, the derivative of the outer function with respect to \(e^{8x}\) is:
\[ -2 \times \frac{1}{\sqrt{1-(e^{8x})^2}} \]
Applying the chain rule:
\[ f'(x) = -2 \times \frac{1}{\sqrt{1-(e^{8x})^2}} \times 8e^{8x} \]
\[ f'(x) = -16 \times \frac{e^{8x}}{\sqrt{1-e^{16x}}} \]
This results in:
\[ f'(x) = -\frac{16e^{8x}}{\sqrt{1-e^{16x}}} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F54e8c682-be67-4ef0-bd1d-cf8b2e543b76%2F5b63e846-8090-45d5-8b86-675e2dc6a1d1%2Fcpuh8hi_processed.png&w=3840&q=75)
Transcribed Image Text:**Mathematical Problem: Finding the Derivative**
Given the function:
\[ f(x) = -2 \sin^{-1} (e^{8x}) \]
We need to find the derivative, \( f'(x) \).
**Solution:**
To find \( f'(x) \), we apply the chain rule. The derivative of \(\sin^{-1}(u)\) with respect to \(u\) is \(\frac{1}{\sqrt{1-u^2}}\).
Given \( u = e^{8x} \), the derivative of \( e^{8x} \) with respect to \(x\) is \( 8e^{8x} \).
Thus, the derivative of the outer function with respect to \(e^{8x}\) is:
\[ -2 \times \frac{1}{\sqrt{1-(e^{8x})^2}} \]
Applying the chain rule:
\[ f'(x) = -2 \times \frac{1}{\sqrt{1-(e^{8x})^2}} \times 8e^{8x} \]
\[ f'(x) = -16 \times \frac{e^{8x}}{\sqrt{1-e^{16x}}} \]
This results in:
\[ f'(x) = -\frac{16e^{8x}}{\sqrt{1-e^{16x}}} \]
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