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Chapter 3: Problem 27

(a) If \(f(x)=\sec x-x,\) find \(f^{\prime}(x)\) . (b) Check to see that your answer to part (a) is reasonable by graphing both \(f\) and \(f^{\prime}\) for \(|x|<\pi / 2\)

Short Answer

Expert verified
(a) \( f'(x) = \sec x \tan x - 1 \). (b) Graph both functions to check. They should correspond on \(|x| < \frac{\pi}{2}\).

Step by step solution

01

Differentiate \( \sec x \)

The function is given as \( f(x) = \sec x - x \). First, we need to find the derivative of \( \sec x \). The derivative of \( \sec x \) is \( \sec x \tan x \).
02

Differentiate \( -x \)

Next, find the derivative of \( -x \). The derivative of \( -x \) is \( -1 \).
03

Combine Derivatives

Now, combine the derivatives from Step 1 and Step 2 using the linearity of differentiation. Hence, \( f'(x) = \sec x \tan x - 1 \).
04

Graph \( f(x) \) and \( f'(x) \)

To verify that our differentiation is correct, graph \( f(x) = \sec x - x \) and \( f'(x) = \sec x \tan x - 1 \) for the interval \(|x| < \frac{\pi}{2}\). The graph of \( f'(x) \) should represent the slope of \( f(x) \) at corresponding points on the interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Secant Function
The secant function, often denoted as \( \sec x \), is one of the basic trigonometric functions. It's defined as the reciprocal of the cosine function. This means \( \sec x = \frac{1}{\cos x} \). Since cosine zeroes at specific intervals, the secant function will have vertical asymptotes, making it undefined at points where cosine equals zero.

Essential characteristics of the secant function include:
  • Periodicity: The function repeats every \(2\pi\).
  • Undefined values: These occur where \(x = \frac{(2n+1)\pi}{2}\) for any integer \(n\).
  • Amplifies when cosine is small, leading to larger values of \( \sec x \).
Visualizing the secant function is crucial for understanding its behavior, especially when combined with other functions, like in the given exercise.
Delving into Differentiation
Differentiation is a fundamental concept in calculus, focusing on finding the derivative of a function. The derivative represents the rate of change or the slope of the function at any given point.

To differentiate \( f(x) = \sec x - x \):
  • Differentiate \( \sec x \): The derivative of \( \sec x \) is \( \sec x \tan x \). This finds how quickly the secant function changes.
  • Differentiate \( -x \): The constant rate of change for \( -x \) is \(-1\). This linearly decreases the function value at a constant rate.
  • Combine the derivatives: Use the rule of linearity in differentiation to get \( f'(x) = \sec x \tan x - 1 \). This combines the rate of change of both parts of the function.
Differentiation allows us to determine not just static values, but dynamic shifts and changes in functions. For trigonometric functions like \( \sec x \), it unveils crucial insights about oscillations and rates of increase or decrease.
Graphing Functions and Their Derivatives
Graphing helps visualize the relationship between a function and its derivative, providing a concrete view of its behavior. For \( f(x) = \sec x - x \) and its derivative \( f'(x) = \sec x \tan x - 1 \), graphing over the interval \(|x| < \frac{\pi}{2}\) offers valuable insights.

Here鈥檚 what to look for while graphing:
  • Function shape: \( f(x) = \sec x - x \) will show oscillatory behavior due to the \( \sec x \) component, trending downwards because of the \(-x\) linear component.
  • Derivative slope: \( f'(x) = \sec x \tan x - 1 \) indicates steepness and direction. Positive values of \( f'(x) \) signify increasing segments of \( f(x) \), while negative values imply decreasing segments.
  • Intercepts and critical points: Points where the derivative crosses zero indicate potential maxima, minima, or inflection points on \( f(x) \).
Graphing both functions is integral to validating the differentiation process. It allows one to visually confirm that the derivative accurately reflects changes in the original function's slope.

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