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
![**Problem Statement:**
Find the derivative of \( y = \cosh^{-1}(8x) \) where \( x > \frac{1}{8}. \)
**Solution:**
To find the derivative of \( y \) with respect to \( x \), we start with the formula for the derivative of the inverse hyperbolic cosine function:
\[
\frac{d}{dx} \cosh^{-1}(u) = \frac{1}{\sqrt{u^2 - 1}}
\]
where \( u = 8x \) in this case.
First, apply the chain rule:
1. Let \( u = 8x \), then \( \frac{du}{dx} = 8 \).
2. Plug \( u = 8x \) into the derivative formula:
\[
\frac{dy}{dx} = \frac{1}{\sqrt{(8x)^2 - 1}} \cdot \frac{du}{dx}
\]
3. Simplify:
\[
\frac{dy}{dx} = \frac{8}{\sqrt{64x^2 - 1}}
\]
So, the derivative \( y' = \frac{8}{\sqrt{64x^2 - 1}} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2432691a-c29a-4a02-b85a-93b716867369%2Fcc7aecd7-2cc1-44d2-9874-ab980157d215%2Fjr5fwcq_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the derivative of \( y = \cosh^{-1}(8x) \) where \( x > \frac{1}{8}. \)
**Solution:**
To find the derivative of \( y \) with respect to \( x \), we start with the formula for the derivative of the inverse hyperbolic cosine function:
\[
\frac{d}{dx} \cosh^{-1}(u) = \frac{1}{\sqrt{u^2 - 1}}
\]
where \( u = 8x \) in this case.
First, apply the chain rule:
1. Let \( u = 8x \), then \( \frac{du}{dx} = 8 \).
2. Plug \( u = 8x \) into the derivative formula:
\[
\frac{dy}{dx} = \frac{1}{\sqrt{(8x)^2 - 1}} \cdot \frac{du}{dx}
\]
3. Simplify:
\[
\frac{dy}{dx} = \frac{8}{\sqrt{64x^2 - 1}}
\]
So, the derivative \( y' = \frac{8}{\sqrt{64x^2 - 1}} \).
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