Question

Find \(f^{\prime}(x)\) for each function. $$ f(x)=\log (\sec x) $$

Short answer

The derivative of the function is \(f'(x) = \tan(x)\).

Step-by-step solution

Understand the Problem

We are given the function \(f(x) = \log(\sec x)\) and need to find its derivative \(f'(x)\). This involves using the chain rule and the identities related to trigonometric and logarithmic functions.

Recall the Derivatives

Recall that the derivative of \(\log(u)\) with respect to \(x\) is \(\frac{1}{u} \cdot \frac{du}{dx}\). Also, the derivative of \(\sec(x)\) is \(\sec(x)\tan(x)\). We will use these derivatives in the next steps.

Apply the Chain Rule

We apply the chain rule to find \(f'(x)\). The outer function is \(\log(u)\) and the inner function \(u = \sec(x)\). The derivative of the outer function is \(\frac{1}{\sec(x)}\), and the derivative of the inner function is \(\sec(x)\tan(x)\).

Compute the Derivative

Using the chain rule, \(f'(x) = \frac{1}{\sec(x)} \cdot \sec(x)\tan(x) = \tan(x)\). This simplifies because the \(\sec(x)\) terms cancel out.

Key concepts

Chain Rule

The chain rule is an essential tool in calculus for finding the derivative of a composite function. A composite function is where you have a function inside another function, like \( f(g(x)) \). In simple terms, the chain rule helps you take derivatives when you need to consider both an inner and an outer function. Here's how it works:

• Identify the inner function \( u = g(x) \) and the outer function as \( f(u) \).
• The chain rule formula is given by \( \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \).

In the exercise, the function \( \log(\sec x) \) can be viewed as having two parts:

• Outer function: \( \log(u) \) where \( u = \sec(x) \)
• Inner function: \( \sec(x) \)

To differentiate \( f(x) = \log(\sec x) \), the chain rule allows us to handle the complexities of having an inner trigonometric function in the logarithmic function. This technique helps ensure an accurate and precise result.

Derivatives of Trigonometric Functions

Trigonometric functions play a vital role in calculus, and understanding their derivatives is fundamental. The main trigonometric functions include \( \sin(x), \cos(x), \tan(x), \csc(x), \sec(x), \) and \( \cot(x) \). Each has its own derivative:

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

In our exercise, when finding the derivative of \( f(x) = \log(\sec x) \), the derivative of the trigonometric function \( \sec(x) \) was needed. We used \( \frac{d}{dx}[\sec(x)] = \sec(x)\tan(x) \) to find the inner part of the differentiation.Knowing these derivatives not only aids calculations but also helps in understanding the relationships and changes in angle measurements.

Logarithmic Differentiation

Logarithmic differentiation is a powerful method used for finding derivatives of functions that are otherwise tough to differentiate. This method uses the properties of logarithms to simplify the differentiation process, especially helpful with products, quotients, or powers.Here's a brief guide on logarithmic differentiation:

• Take the natural logarithm of both sides of the equation: \( y = f(x) \rightarrow \ln(y) = \ln(f(x)) \).
• Use the property of logarithms to simplify: \( \ln(abc) = \ln(a) + \ln(b) + \ln(c) \).
• Differentiate implicitly with respect to \( x \).
• Solve for \( y' \), and substitute back \( y = f(x) \).

In the provided exercise, we didn't use logarithmic differentiation in the traditional sense because the function was \( \log(\sec x) \), but understanding how to differentiate \( \log(x) \) functions is part of why this method is effective. The derivative of \( \log(x) \) itself is given by \( \frac{1}{x} \), a key step in solving log-related problems. Familiarity with these processes helps simplify otherwise complex differentiation tasks.