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 \(\frac{dy}{dx}\), where:
\[ y = 3(\sec x + \tan x)(\sec x - \tan x) \]
---
**Solution:**
To find \(\frac{dy}{dx}\), we will need to first expand the expression for \(y\):
\[ y = 3[(\sec x + \tan x)(\sec x - \tan x)] \]
Using the identity \((a+b)(a-b) = a^2 - b^2\), we have:
\[ y = 3(\sec^2 x - \tan^2 x) \]
Next, we'll use the identity \(\sec^2 x - \tan^2 x = 1\):
\[ y = 3 \times 1 = 3 \]
Now, since \(y\) is a constant, the derivative \(\frac{dy}{dx} = 0\).
Therefore, \(\frac{dy}{dx} = \boxed{0}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F256dbc8b-2bc9-463c-8f55-d3702041f8b6%2F263cceb3-d1c0-4ba2-af1d-2ddecc0ef304%2Fht2k6dq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find \(\frac{dy}{dx}\), where:
\[ y = 3(\sec x + \tan x)(\sec x - \tan x) \]
---
**Solution:**
To find \(\frac{dy}{dx}\), we will need to first expand the expression for \(y\):
\[ y = 3[(\sec x + \tan x)(\sec x - \tan x)] \]
Using the identity \((a+b)(a-b) = a^2 - b^2\), we have:
\[ y = 3(\sec^2 x - \tan^2 x) \]
Next, we'll use the identity \(\sec^2 x - \tan^2 x = 1\):
\[ y = 3 \times 1 = 3 \]
Now, since \(y\) is a constant, the derivative \(\frac{dy}{dx} = 0\).
Therefore, \(\frac{dy}{dx} = \boxed{0}\).
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