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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![**Topic: Calculus - Derivatives of Inverse Hyperbolic Functions**
**Exercise: Find the Derivative**
Given the function:
\[ y = \sinh^{-1} \left( \frac{x}{2} \right) \]
Determine the derivative \(\frac{dy}{dx}\).
**Solution:**
To find the derivative, we utilize the formula for the derivative of the inverse hyperbolic sine function:
For \( y = \sinh^{-1}(u) \), the derivative is \(\frac{dy}{dx} = \frac{1}{\sqrt{u^2 + 1}} \cdot \frac{du}{dx}\).
In this exercise, \( u = \frac{x}{2} \).
1. Calculate \(\frac{du}{dx}\):
\[\frac{du}{dx} = \frac{1}{2}\]
2. Substitute \( u = \frac{x}{2} \) into the derivative formula:
\[\frac{dy}{dx} = \frac{1}{\sqrt{\left(\frac{x}{2}\right)^2 + 1}} \cdot \frac{1}{2}\]
3. Simplify:
\[\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{x^2}{4} + 1}} = \frac{1}{2\sqrt{\frac{x^2 + 4}{4}}} = \frac{1}{\sqrt{x^2 + 4}}\]
Thus, the derivative is:
\[\frac{dy}{dx} = \frac{1}{\sqrt{x^2 + 4}}\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2e33d8a8-4a77-428f-baec-aa6f37a9e0af%2F0eec6bfb-f288-4536-890d-1a4c72682448%2F0fkk3yq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Topic: Calculus - Derivatives of Inverse Hyperbolic Functions**
**Exercise: Find the Derivative**
Given the function:
\[ y = \sinh^{-1} \left( \frac{x}{2} \right) \]
Determine the derivative \(\frac{dy}{dx}\).
**Solution:**
To find the derivative, we utilize the formula for the derivative of the inverse hyperbolic sine function:
For \( y = \sinh^{-1}(u) \), the derivative is \(\frac{dy}{dx} = \frac{1}{\sqrt{u^2 + 1}} \cdot \frac{du}{dx}\).
In this exercise, \( u = \frac{x}{2} \).
1. Calculate \(\frac{du}{dx}\):
\[\frac{du}{dx} = \frac{1}{2}\]
2. Substitute \( u = \frac{x}{2} \) into the derivative formula:
\[\frac{dy}{dx} = \frac{1}{\sqrt{\left(\frac{x}{2}\right)^2 + 1}} \cdot \frac{1}{2}\]
3. Simplify:
\[\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{x^2}{4} + 1}} = \frac{1}{2\sqrt{\frac{x^2 + 4}{4}}} = \frac{1}{\sqrt{x^2 + 4}}\]
Thus, the derivative is:
\[\frac{dy}{dx} = \frac{1}{\sqrt{x^2 + 4}}\]
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