Question

Let \(f(x)=x[\sin x-\cos (x / 2)]\). (a) Draw the graphs of \(f(x), f^{\prime}(x), f^{\prime \prime}(x)\), and \(f^{\prime \prime \prime}(x)\) on \([0,6]\) using the same axes. (b) Evaluate \(f^{\prime \prime \prime}(2.13)\).

Short answer

Calculate \(f'''(2.13)\) using a calculator.

Step-by-step solution

Differentiate the Function f(x) to Find f'(x)

To find the first derivative of the function, apply the product rule. Given the function \[f(x) = x \sin(x) - x \cos\left(\frac{x}{2}\right)\]The first derivative is calculated by applying the product rule to each term:For the first term: \(g(x) = x, \, h(x) = \sin(x)\),\[f_1'(x) = x \cos(x) + \sin(x)\]For the second term: \(g(x) = x, \, h(x) = \cos\left(\frac{x}{2}\right)\),\[f_2'(x) = - x \cdot \frac{1}{2} \sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\]Thus,\[f'(x) = x \cos(x) + \sin(x) + \cos\left(\frac{x}{2}\right) - \frac{x}{2} \sin\left(\frac{x}{2}\right)\]

Differentiate f'(x) to Find f''(x)

Take the derivative of the first derivative function to find the second derivative:\[f'(x) = x\cos(x) + \sin(x) + \cos\left(\frac{x}{2}\right) - \frac{x}{2} \sin\left(\frac{x}{2}\right)\]Differentiate each term separately:\[f''(x) = \frac{d}{dx}(x\cos(x)) + \frac{d}{dx}(\sin(x)) + \frac{d}{dx}\left(\cos\left(\frac{x}{2}\right)\right) - \frac{d}{dx}\left(\frac{x}{2} \sin\left(\frac{x}{2}\right)\right)\]Calculate each derivative:1. \(\frac{d}{dx}(x\cos(x)) = -x\sin(x) + \cos(x)\)2. \(\frac{d}{dx}(\sin(x)) = \cos(x)\)3. \(\frac{d}{dx}\left(\cos\left(\frac{x}{2}\right)\right) = -\frac{1}{2}\sin\left(\frac{x}{2}\right)\)4. \(\frac{d}{dx}\left(\frac{x}{2}\sin\left(\frac{x}{2}\right)\right) = \frac{1}{2}\sin\left(\frac{x}{2}\right) + \frac{x}{4}\cos\left(\frac{x}{2}\right)\)Combine all terms:\[f''(x) = \cos(x) + \cos(x) - x\sin(x) - \frac{1}{2}\sin\left(\frac{x}{2}\right) - \frac{1}{2}\sin\left(\frac{x}{2}\right) - \frac{x}{4}\cos\left(\frac{x}{2}\right)\]Simplify it to:\[f''(x) = 2\cos(x) - x\sin(x) - \sin\left(\frac{x}{2}\right) - \frac{x}{4}\cos\left(\frac{x}{2}\right)\]

Differentiate f''(x) to Find f'''(x)

Differentiate the second derivative to get the third derivative:\[f''(x) = 2\cos(x) - x\sin(x) - \sin\left(\frac{x}{2}\right) - \frac{x}{4}\cos\left(\frac{x}{2}\right)\]Calculate each derivative:1. \(\frac{d}{dx}(2\cos(x)) = -2\sin(x)\)2. \(\frac{d}{dx}(-x\sin(x)) = -x\cos(x) - \sin(x)\)3. \(\frac{d}{dx}(-\sin\left(\frac{x}{2}\right)) = -\frac{1}{2}\cos\left(\frac{x}{2}\right)\)4. \(\frac{d}{dx}\left(- \frac{x}{4}\cos\left(\frac{x}{2}\right)\right) = -\frac{1}{4}\cos\left(\frac{x}{2}\right) + \frac{x}{8} \sin\left(\frac{x}{2}\right)\)Combine all terms:\[f'''(x) = -2\sin(x) - x\cos(x) - \sin(x) - \frac{1}{2}\cos\left(\frac{x}{2}\right) - \frac{1}{4}\cos\left(\frac{x}{2}\right) + \frac{x}{8}\sin\left(\frac{x}{2}\right)\]Simplify it to:\[f'''(x) = -3\sin(x) - x\cos(x) - \frac{3}{4}\cos\left(\frac{x}{2}\right) + \frac{x}{8}\sin\left(\frac{x}{2}\right)\]

Evaluate f'''(2.13)

Now compute \(f'''(2.13)\) using the formula:\[f'''(x) = -3\sin(x) - x\cos(x) - \frac{3}{4}\cos\left(\frac{x}{2}\right) + \frac{x}{8}\sin\left(\frac{x}{2}\right)\]Substitute 2.13 into the equation and perform the calculations:\[f'''(2.13) = -3\sin(2.13) - 2.13\cos(2.13) - \frac{3}{4}\cos\left(\frac{2.13}{2}\right) + \frac{2.13}{8}\sin\left(\frac{2.13}{2}\right)\]Compute numerical values using a calculator for accurate results. This will give you the value for \(f'''(2.13)\).

Key concepts

Derivative

A derivative helps us understand how a function changes. It's like finding the speed of a car at any moment if you know how its position changes over time. The derivative, denoted as \( f'(x) \), is a technique in calculus to find the slope of a curve at any given point. For a given function \( f(x) \), the derivative tells us how \( f(x) \) changes as \( x \) changes. When differentiating, you often use rules like the product rule, quotient rule, or chain rule depending on the function type. In the exercise, we apply the product rule to differentiate the functions, which states that if you have a product of two functions, \( u(x)v(x) \), its derivative is \( u'(x)v(x) + u(x)v'(x) \). With the derivative formula, you can determine both the instantaneous rate of change of a variable and the function's critical points, maxima, and minima.

Trigonometric Functions

Trigonometric functions, such as \( \sin(x) \) and \( \cos(x) \), are periodic functions that repeat their values in regular intervals. These functions are essential in calculus for describing oscillations, waves, and various other phenomena in engineering and physics. The function \( \sin(x) \) varies between -1 and 1, as does \( \cos(x) \), and both have unique properties and graphs. In derivatives involving trigonometric functions, certain rules like the derivatives of \( \sin(x) = \cos(x) \) and \( \cos(x) = -\sin(x) \) are crucial. These properties are used to simplify and calculate derivatives precisely, as seen in our step-by-step solution where we handle expressions like \( x \sin(x) \) and \( \cos(\frac{x}{2}) \). Understanding how to differentiate trigonometric functions can lead to deeper insights into their behavior, such as their rates of change at various points.

Graphing Functions

Graphing functions allows us to visualize how they behave over an interval, such as \([0,6]\) in our exercise. This visual interaction helps in identifying points of interest like peaks, troughs, and inflection points without calculation. When graphing \( f(x) \) and its derivatives \( f'(x), f''(x), \) and \( f'''(x) \), it's essential to observe the impacts of changes. For instance, the first derivative graph shows where the function is increasing or decreasing. The second derivative hints at the curvature or concavity of the graph, while the third derivative graph highlights changes in concavity. Visualizing these aspects with actual graphs makes it easier to understand and predict the behavior of functions across different domains and ranges. Graphs provide a powerful tool for both interpreting and making predictions about real-world phenomena modeled by the functions.

Higher-Order Derivatives

Higher-order derivatives tell us more about the change in a function beyond the basic rate of change. They involve taking the derivative of a derivative. So, for a function \( f(x) \), its first derivative \( f'(x) \) explains how \( f(x) \) changes, the second derivative \( f''(x) \) indicates how the rate of change itself changes, and the third derivative \( f'''(x) \) gives us even deeper insights.

• The second derivative \( f''(x) \) helps in understanding the concavity of the function. If \( f''(x) > 0 \), the graph is concave up like a cup. If \( f''(x) < 0 \), it's concave down like a frown.
• The third derivative \( f'''(x) \) can reveal points of inflection where the concavity of the function changes.

In our exercise, we find \( f'''(2.13) \) to analyze the rate at which the curvature changes at that point. Calculating higher-order derivatives can be complex, but by using rules systematically, you can extract meaningful insights about intricate function behaviors.