Problem 24 - Calculus Single Variable

Question

An expression for \(f(x)\) is given. Compute the first, second, and third derivatives of \(f(x)\) with respect to \(x\). \(2^{x} \cos (3 x)\)

Step-by-Step Solution

Answer

The derivatives of \( f(x) = 2^x \cos(3x) \) are: \( f'(x) = 2^x(\ln(2)\cos(3x) - 3\sin(3x)) \), \( f''(x) = 2^x(\ln^2(2)\cos(3x) - 6\ln(2)\sin(3x) - 9\cos(3x)) \), \( f'''(x) = 2^x(\ln^3(2)\cos(3x) - 12\ln^2(2)\sin(3x) + 27\sin(3x) - 9\ln(2)\cos(3x)) \).

Step 1: Calculate the First Derivative

The given function is \( f(x) = 2^x \cos(3x) \). To find the derivative of \( f(x) \), we need to use the product rule, which states that if \( u(x) = 2^x \) and \( v(x) = \cos(3x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \). 1. Compute \( u'(x) \):\[ u(x) = 2^x \Rightarrow u'(x) = 2^x \ln(2) \].2. Compute \( v'(x) \):\[ v(x) = \cos(3x) \Rightarrow v'(x) = -3 \sin(3x) \].Plug these into the product rule:\[ f'(x) = 2^x \ln(2) \cos(3x) + 2^x (-3 \sin(3x)) \].This simplifies to \[ f'(x) = 2^x (\ln(2) \cos(3x) - 3 \sin(3x)) \].

Step 2: Calculate the Second Derivative

To find the second derivative \( f''(x) \), take the derivative of the first derivative \( f'(x) \). Use the product rule again:Let \( u(x) = 2^x \) and derive \( f'(x) = 2^x (\ln(2) \cos(3x) - 3 \sin(3x)) \). We find:1. Derive \( 2^x \) again as \( 2^x \ln(2) \).2. Use the product rule:\[ f''(x) = \left(2^x \ln(2)\right)(\ln(2) \cos(3x) - 3 \sin(3x)) + 2^x (-3) (\ln(2) \sin(3x) + 3 \cos(3x)) \].3. Simplify: \[ f''(x) = 2^x \left( \ln^2(2) \cos(3x) - 3 \ln(2) \sin(3x) - 3 \ln(2) \sin(3x) - 9 \cos(3x) \right) \].

Step 3: Calculate the Third Derivative

The third derivative \( f'''(x) \) requires deriving \( f''(x) \). We need to simplify and apply the product rule one last time:\[ f'''(x) = \text{Derive of } 2^x (\ln^2(2) \cos(3x) - 6 \ln(2) \sin(3x) - 9 \cos(3x)) \].Using similar steps as above:1. \( g(x) = 2^x \) gives \( g'(x) = 2^x \ln(2) \).2. Apply the product rule for each part separately and combine terms:\[ f'''(x) = 2^x \left( \ln^3(2) \cos(3x) - 9 \ln(2) \cos(3x) - 12 \ln^2(2) \sin(3x) + 27 \sin(3x) \right) \].

Step 4: Summarize the Derivatives

The first three derivatives of \( f(x) = 2^x \cos(3x) \) calculated are:1. First Derivative: \[ f'(x) = 2^x (\ln(2) \cos(3x) - 3 \sin(3x)) \]2. Second Derivative: \[ f''(x) = 2^x (\ln^2(2) \cos(3x) - 6 \ln(2) \sin(3x) - 9 \cos(3x)) \]3. Third Derivative: \[ f'''(x) = 2^x (\ln^3(2) \cos(3x) - 12 \ln^2(2) \sin(3x) + 27 \sin(3x) - 9 \ln(2) \cos(3x)) \]

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus used to differentiate functions that are the product of two other functions. Understanding this rule is essential for solving derivatives when functions are multiplied. If you have a function written as a product, like \( u(x) \cdot v(x) \), where \( u(x) \) and \( v(x) \) are two separate functions, the product rule says that:

• \( (u \cdot v)' = u'v + uv' \)

This means that to find the derivative of the product, you differentiate the first function \( u(x) \), multiply it by the second function \( v(x) \) left untouched, then add the product of \( u(x) \) and the derivative of the second function \( v(x) \).
This rule ensures that every possible rate of change within the function is accounted for.
In the context of our problem, where \( u(x) = 2^x \) and \( v(x) = \cos(3x) \), applying the product rule helps compute each derivative step.

First Derivative

The first derivative represents the rate of change of the function, or its slope, at any given point. For a given function \( f(x) \), you obtain the first derivative \( f'(x) \) through application of differentiation rules.
In our exercise, we apply the product rule to \( f(x) = 2^x \cos(3x) \):

• First, \( u'(x) = 2^x \ln(2) \) since the exponential derivative \( a^x \) results in \( a^x \ln(a) \).
• Next, \( v'(x) = -3\sin(3x) \) due to the chain rule applied to \( \cos(3x) \).

Putting these into the product rule:

• \( f'(x) = 2^x \ln(2) \cos(3x) + 2^x (-3 \sin(3x)) \)

Finally, simplifying:

• \( f'(x) = 2^x (\ln(2) \cos(3x) - 3 \sin(3x)) \)

This derivative tells us how the function \( f(x) \) behaves as \( x \) changes, capturing its instantaneous rate of change.

Second Derivative

The second derivative, denoted as \( f''(x) \), looks at the rate of change of the first derivative, offering insights into the function's concavity.
Essentially, the second derivative helps us understand how the rate of change itself is changing, indicating if the function is curving upwards or downwards.

• For our function, reapply the product rule to \( f'(x) = 2^x (\ln(2) \cos(3x) - 3 \sin(3x)) \).

This involves:

• Derivative of \( 2^x \) again, which is \( 2^x \ln(2) \).

Breaking it down:

• Off parts combine as: \( f''(x) = \{2^x \ln(2)\} \cdot \{\ln(2) \cos(3x) - 3 \sin(3x)\} + 2^x \cdot \{-3 (\ln(2) \sin(3x) + 3 \cos(3x))\} \)

After simplification, we find:

• \( f''(x) = 2^x (\ln^2(2) \cos(3x) - 6 \ln(2) \sin(3x) - 9 \cos(3x)) \)

This provides a picture of the acceleration of the function, i.e., how quickly the slope is changing.

Third Derivative

The third derivative, \( f'''(x) \), explores how the second derivative is changing, giving us information about the jerk or the rate of change of acceleration of the original function.
It delves even deeper into understanding the function's behavior, especially when analyzing dynamic systems.

• Returning to our second derivative, it's crucial to apply the product rule yet again.

• Using what we've obtained previously: \( f''(x) = 2^x (\ln^2(2) \cos(3x) - 6 \ln(2) \sin(3x) - 9 \cos(3x)) \)

Computing \( f'''(x) \):

• Focus initially on \( 2^x \ln(2) \) and apply the rule for complex internal terms.

Simplified into:

• \( f'''(x) = 2^x (\ln^3(2) \cos(3x) - 12 \ln^2(2) \sin(3x) + 27 \sin(3x) - 9 \ln(2) \cos(3x)) \)

This derivative is valuable in technical scenarios involving rates of change upon rates of change, providing a sophisticated understanding of the underlying function's variability.