Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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**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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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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