Problem 15 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$y=(\sec x+\tan x)(\sec x-\tan x)$$

Step-by-Step Solution

Answer

The derivative \(dy/dx = 0\).

Step 1: Identify the Expression to Differentiate

We need to differentiate the expression \[y = (\sec x + \tan x)(\sec x - \tan x)\]This is a product of two binomials.

Step 2: Use Product-to-Sum Formulas

Recognize that \[(\sec x + \ an x)(\sec x - \ an x) = \sec^2x - \tan^2x\]This formula works similarly to the difference of squares: \[a^2 - b^2 = (a+b)(a-b)\]

Step 3: Simplify the Expression

Now simplify the expression \[\sec^2x - \tan^2x\].Utilize the Pythagorean identity: \[1 + \tan^2x = \sec^2x\] to simplify as follows:\[\sec^2x - \tan^2x = 1\].

Step 4: Differentiate the Simplified Expression

Now that the expression simplifies to 1, differentiate it with respect to \(x\):\[\frac{d}{dx}(1) = 0\].

Key Concepts

Difference of Squares

The difference of squares is a powerful algebraic tool that helps simplify expressions involving a product of binomials. When you have two terms, say \(a\) and \(b\), the difference of squares formula states that:

• \(a^2 - b^2 = (a+b)(a-b)\)

This means that if you recognize a product of binomials in the form \((a+b)(a-b)\), you can instantly simplify it to the difference of squares, \(a^2 - b^2\).
In this particular exercise, you've observed an expression \((\sec x + \tan x)(\sec x - \tan x)\). By identifying it as a difference of squares, it simplifies to \(\sec^2 x - \tan^2 x\). This makes it much easier to work with, especially for further steps such as differentiation. Keeping an eye out for these patterns can greatly optimize your process in calculus and algebra.

Pythagorean Identity

The Pythagorean identity is a fundamental rule in trigonometry, often used to simplify expressions involving trigonometric functions. One of the most common versions of this identity is:

• \(1 + \tan^2x = \sec^2x\)

Recognizing and applying this identity allows us to simplify complex trigonometric expressions to more manageable forms.
In the original exercise, after applying the difference of squares to get \(\sec^2 x - \tan^2 x\), the Pythagorean identity is used strategically. Substituting the identity directly helps in simplifying the expression further. Knowing that \(1 + \tan^2 x = \sec^2 x\), we rewrite \(\sec^2 x - \tan^2 x\) to \(1\). This step is crucial before you proceed with differentiation, as it greatly simplifies the problem.

Trigonometric Functions

Trigonometric functions like secant \(\sec x\) and tangent \(\tan x\) are essential components of calculus, especially when dealing with expressions and their derivatives. Each trigonometric function has its derivatives:

• The derivative of \(\sec x\) is \(\sec x \tan x\)
• The derivative of \(\tan x\) is \(\sec^2 x\)

Understanding how these functions behave and relate to one another is important for solving calculus problems.
In the given exercise, although the differentiation ended up being quite straightforward because the expression simplified to a constant (1), it's still valuable to note how we simplify using these identities and functions. By simplifying \((\sec x + \tan x)(\sec x - \tan x)\) using the difference of squares and then employing the Pythagorean identity, we reduce any trigonometric complexity. This approach ensures accuracy and efficiency when dealing with derivatives of trigonometric functions in calculus.