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
![**Differentiating the Function**
Given the function:
\[ y = \sqrt{\sec x + 2021} \]
To find the derivative of this function with respect to \( x \), we will apply the chain rule. The chain rule is used when differentiating composite functions. Here, the outer function is the square root, and the inner function is \(\sec x + 2021\).
**Steps to Differentiate:**
1. **Identify the outer function and its derivative:**
- The outer function is \(\sqrt{u}\), where \(u = \sec x + 2021\).
- The derivative of \(\sqrt{u}\) with respect to \(u\) is \(\frac{1}{2\sqrt{u}}\).
2. **Identify the inner function and its derivative:**
- The inner function is \(u = \sec x + 2021\).
- The derivative of \(\sec x\) with respect to \(x\) is \(\sec x \tan x\).
- The constant 2021 has a derivative of 0.
3. **Apply the chain rule:**
- The derivative of the function \(y\) is given by:
\[
\frac{dy}{dx} = \frac{1}{2\sqrt{\sec x + 2021}} \cdot (\sec x \tan x)
\]
Therefore, the derivative of the function is:
\[
\frac{dy}{dx} = \frac{\sec x \tan x}{2\sqrt{\sec x + 2021}}
\]
This derivative gives the rate of change of the function with respect to \(x\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff80585f9-eac0-4aa8-81f9-ee704a35044a%2Fd64304d1-4ab5-4005-8ef0-600a03284158%2F6sa2wmhv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Differentiating the Function**
Given the function:
\[ y = \sqrt{\sec x + 2021} \]
To find the derivative of this function with respect to \( x \), we will apply the chain rule. The chain rule is used when differentiating composite functions. Here, the outer function is the square root, and the inner function is \(\sec x + 2021\).
**Steps to Differentiate:**
1. **Identify the outer function and its derivative:**
- The outer function is \(\sqrt{u}\), where \(u = \sec x + 2021\).
- The derivative of \(\sqrt{u}\) with respect to \(u\) is \(\frac{1}{2\sqrt{u}}\).
2. **Identify the inner function and its derivative:**
- The inner function is \(u = \sec x + 2021\).
- The derivative of \(\sec x\) with respect to \(x\) is \(\sec x \tan x\).
- The constant 2021 has a derivative of 0.
3. **Apply the chain rule:**
- The derivative of the function \(y\) is given by:
\[
\frac{dy}{dx} = \frac{1}{2\sqrt{\sec x + 2021}} \cdot (\sec x \tan x)
\]
Therefore, the derivative of the function is:
\[
\frac{dy}{dx} = \frac{\sec x \tan x}{2\sqrt{\sec x + 2021}}
\]
This derivative gives the rate of change of the function with respect to \(x\).
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