Problem 334 - Calculus Volume 1

Question

For the following exercises, evaluate by any method. $$\frac{d}{d x} \ln (\sec x+\tan x)$$

Step-by-Step Solution

Answer

The derivative is \( \sec x \).

Step 1: Recognize the Derivative Rule

The given function to differentiate is \( \ln(\sec x + \tan x) \). We need to use the derivative of a natural logarithm function which is \( \frac{d}{dx}[\ln u] = \frac{1}{u} \cdot \frac{du}{dx} \), where \( u \) is the inside function \( \sec x + \tan x \) in this case.

Step 2: Differentiate the Inner Function

Now, differentiate the inner function \( u = \sec x + \tan x \). The derivatives are \( \frac{d}{dx}[\sec x] = \sec x \tan x \) and \( \frac{d}{dx}[\tan x] = \sec^2 x \). Therefore, \( \frac{du}{dx} = \sec x \tan x + \sec^2 x \).

Step 3: Apply the Chain Rule

Using the derivative formula from Step 1, apply the chain rule: \( \frac{d}{dx}[\ln(\sec x + \tan x)] = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x) \).

Step 4: Simplify the Expression

The expression becomes \( \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} \). Since both the numerator and denominator contain similar terms, the expression simplifies directly to \( \sec x \). This is because the terms divide out cleanly given that numerator \( \sec x \tan x + \sec^2 x \) equals the product form of the terms in the denominator when distributed.

Key Concepts

Chain Rule in Calculus

The chain rule is a fundamental concept in calculus that allows us to differentiate composite functions. This means we can find the derivative of a function that is composed of another function.
For example, if we have a function that is the composition of two functions, like \( f(g(x)) \), the chain rule helps us differentiate it efficiently.
The formula for the chain rule is:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).

In our exercise, the function \( \ln(\sec x + \tan x) \) can be broken down into an outer function, \( \ln(u) \), and an inner function, \( u = \sec x + \tan x \).
By applying the chain rule, we first find the derivative of the outer function in terms of the inner function, \( \frac{d}{du}[\ln u] = \frac{1}{u} \). Furthermore, we multiply it by the derivative of the inner function, \( \frac{du}{dx} \). This is crucial for analyzing complex functions, ensuring all components are aligned correctly.

Differentiation Techniques

Differentiation involves various techniques to handle different types of functions, making it a powerful tool in calculus.
Some techniques you might encounter include:

• The power rule for polynomials.
• The product rule for functions multiplied together.
• The quotient rule for functions divided by each other.
• The chain rule, which we already discussed.

In the given problem, the focus is on differentiating using the **natural logarithmic function**. The derivative of \( \ln u \), where \( u \) is a function of \( x \), is given by \( \frac{1}{u} \cdot \frac{du}{dx} \). This is a direct application of the chain rule here, which is essential for handling logarithmic functions combined with other types.
The differentiation technique used here simplifies complex expressions and aids in solving real-world problems by providing the slope or rate of change of functions. By mastering these techniques, you can tackle calculus problems across various fields.

Trigonometric Derivatives

Trigonometric derivatives are particularly useful when dealing with functions involving sine, cosine, tangent and their reciprocals like secant.
Some essential trigonometric derivatives are:

• \( \frac{d}{dx}[\sin x] = \cos x \)
• \( \frac{d}{dx}[\cos x] = -\sin x \)
• \( \frac{d}{dx}[\tan x] = \sec^2 x \)
• \( \frac{d}{dx}[\sec x] = \sec x \tan x \)

In the exercise, the function \( \sec x + \tan x \) was differentiated using these derivatives, giving us \( \sec x \tan x + \sec^2 x \).
Recognizing these trigonometric derivatives allows us to easily approach more complex problems and streamline our calculations.
Whether it's physics, engineering, or even economics, trigonometric derivatives provide the ability to model periodic behaviors and solve dynamic problems.