Problem 26 - Thomas Calculus

Question

In Exercises \(5-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\ln (\sec \theta+\tan \theta) $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{d\theta} = \frac{\sec \theta \tan \theta + \sec^2 \theta}{\sec \theta + \tan \theta} \).

Step 1: Identify the Function

We are given the function \( y = \ln (\sec \theta + \tan \theta) \). Our task is to find its derivative with respect to \( \theta \).

Step 2: Use the Chain Rule

To differentiate \( y = \ln (\sec \theta + \tan \theta) \), we apply the chain rule. This means we need to find the derivative of the inner function \( u = \sec \theta + \tan \theta \) first, then use that to find the derivative of \( \ln(u) \).

Step 3: Differentiate the Inner Function

Differentiate \( u = \sec \theta + \tan \theta \) with respect to \( \theta \).- The derivative of \( \sec \theta \) is \( \sec \theta \tan \theta \).- The derivative of \( \tan \theta \) is \( \sec^2 \theta \).Thus, \( \frac{du}{d\theta} = \sec \theta \tan \theta + \sec^2 \theta \).

Step 4: Differentiate the Logarithmic Function

Now use the derivative of the logarithmic function \( \frac{d}{du} \ln(u) = \frac{1}{u} \). So, the derivative of \( y = \ln(u) \) is \( \frac{dy}{d\theta} = \frac{1}{u} \cdot \frac{du}{d\theta} \).

Step 5: Substitute and Simplify

Substitute \( u = \sec \theta + \tan \theta \) and \( \frac{du}{d\theta} = \sec \theta \tan \theta + \sec^2 \theta \) into the derivative expression. Thus,\[ \frac{dy}{d\theta} = \frac{1}{\sec \theta + \tan \theta} \cdot (\sec \theta \tan \theta + \sec^2 \theta) \].

Key Concepts

Chain Rule

The chain rule is a powerful tool in calculus used to differentiate composite functions. Composite functions are functions made up of two or more functions. In simple terms, the chain rule allows us to find the derivative of a function that is nested within another function. For example, when you have a function of the form \( y = f(g(x)) \), the chain rule helps you break it into manageable parts.

When applying the chain rule, we first differentiate the outer function and then multiply it by the derivative of the inner function. This method works well for finding derivatives of complicated functions by breaking them down into simpler components.

• If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

Using the chain rule simplifies the problem and paves the way to tackling derivatives of more complex expressions.

Logarithmic Differentiation

Logarithmic differentiation is especially useful when dealing with functions that are products or quotients of other functions, or where the function involves a logarithm, as in our exercise. It allows simplification by converting multiplicative relationships into additive ones through the use of logarithms.

Here's the core idea:

• Take the natural logarithm of both sides of the equation, \( y = \, \text{your function} \).
• Use properties of logarithms to simplify.
• Differentiate the resulting equation using known derivatives and rules.

This technique not only helps in managing complex expressions but also in reducing computation errors that often arise from direct differentiation of difficult functions.
In the exercise, the use of logarithmic differentiation helped with differentiating the function \( y = \ln(\sec \theta + \tan \theta) \), allowing us to efficiently manage the composite function inside the log.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are foundational in calculus and mathematical analysis. Understanding how these functions are differentiated is crucial.

The main trigonometric functions and their derivatives include:

• If \( y = \sin(x) \), then \( \frac{dy}{dx} = \cos(x) \).
• If \( y = \cos(x) \), then \( \frac{dy}{dx} = -\sin(x) \).
• If \( y = \tan(x) \), then \( \frac{dy}{dx} = \sec^2(x) \).
• If \( y = \sec(x) \), then \( \frac{dy}{dx} = \sec(x)\tan(x) \).

In our example, the derivative of the trigonometric functions \( \sec \theta \) and \( \tan \theta \) were used. Transforming these functions into their derivatives helps in constructing the derivative of more complex trigonometric-based expressions, making them manageable to solve through differentiation methods like the chain rule.