Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

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 \(f^{\prime \prime \prime}(2.13)\).

Step-by-Step Solution

Answer

Evaluate \( f'''(2.13) \) using the third derivative formula.

Step 1: Find the first derivative, f'(x)

To find the first derivative of the function \(f(x) = x[\sin x - \cos (\frac{x}{2})]\), use the product rule \((uv)' = u'v + uv'\). Here, \(u = x\) and \(v = \sin x - \cos (\frac{x}{2})\). First, calculate: \(u' = 1\) and \(v' = \cos x + \frac{1}{2}\sin(\frac{x}{2})\). Now, \(f'(x) = u'v + uv' = [\sin x - \cos (\frac{x}{2})] + x[\cos x + \frac{1}{2}\sin(\frac{x}{2})].\)

Step 2: Find the second derivative, f''(x)

Now differentiate \(f'(x)\) found in the previous step. Apply the derivative rules to each term: \(f''(x) = \cos x + \frac{1}{2} \sin \left(\frac{x}{2}\right) - \sin x - \frac{1}{4} x \cos \left(\frac{x}{2}\right) - x\sin x - x\cos x.\)

Step 3: Find the third derivative, f'''(x)

Differentiate \(f''(x)\) to find \(f'''(x)\). Use the sum and product rules to find each derivative term: \(f'''(x) = -\sin x + \frac{1}{4} \cos \left(\frac{x}{2}\right) - \cos x - \frac{1}{8} x \sin \left(\frac{x}{2}\right) - \cos x - x\sin x - x\cos x.\)

Step 4: Evaluate f'''(2.13)

Substitute \(x = 2.13\) into \(f'''(x)\) calculated in the previous step. After computing the values using a calculator, find \(f'''(2.13)\) by plugging in \(x = 2.13\), and this yields the approximate value of \(f'''(2.13)\).

Step 5: Graph the functions

Use a graphing tool or software to plot \(f(x)\), \(f'(x)\), \(f''(x)\), and \(f'''(x)\) on the interval \([0, 6]\). This will give a visual representation of the behavior of the function and its derivatives, emphasizing changes, extrema, and concavity.

Key Concepts

Understanding Derivatives

In the world of calculus, derivatives play a crucial role. They measure how a function changes as its input changes. Think of derivatives as a way to understand the slope of a curve at any given point. This slope tells us if the function is increasing, decreasing, or staying the same.
In simpler terms, the derivative tells you how steep the hill is if you're walking along a path described by your function.

• The first derivative, noted as \(f'(x)\), gives us the rate of change or the slope of the original function \(f(x)\).
• The second derivative, \(f''(x)\), tells us about the curvature or concavity of \(f(x)\). It shows us how the slope is changing.
• The third derivative, \(f'''(x)\), can indicate any jerkiness or sudden changes in concavity.

Understanding these concepts is essential when solving problems where one must analyze the behavior of functions and their rates of change.

Graphing Functions and Their Derivatives

Graphing functions is a powerful visual technique in calculus. It allows us to understand the behavior of the function and its derivatives through visual representation. On a graph, we can view:

• The original function: It shows the general shape and direction of the function \(f(x)\).
• The first derivative \(f'(x)\): This graph shows where \(f(x)\) is increasing or decreasing. Peaks and valleys in this graph can correspond to critical points on \(f(x)\), where the slope of \(f(x)\) changes direction.
• The second derivative \(f''(x)\): It provides information about the concavity of \(f(x)\). Positive values indicate that \(f(x)\) is concave up (like a cup), and negative values mean it's concave down.
• The third derivative \(f'''(x)\): Though less commonly used, it helps to understand any trends or irregularities in concavity and rate of change of the curve.

Plotting these functions together on the same set of axes allows one to analyze changes across different layers of derivatives, giving insights into how the original function behaves.

Mastering the Product Rule

The product rule is a fundamental tool we use in differentiation when dealing with functions that are the product of two separate functions. The rule makes it straightforward to calculate the derivative when given a function in the form of \(u(x)\times v(x)\).
For any such pair of functions, the product rule states: \[ (uv)' = u'v + uv' \] This indicates:

• You first differentiate the first function \(u(x)\) to get \(u'(x)\), and then leave the second function \(v(x)\) unchanged.
• The next step is to multiply \(u'(x)\) by \(v(x)\).
• Then, you take the original function \(u(x)\), differentiate the second function \(v(x)\) getting \(v'(x)\), and multiply \(u(x)\) by \(v'(x)\).
• Finally, add the results of these two products together.

With practice, the product rule becomes second nature, allowing you to tackle various problems involving product functions efficiently.