Problem 23 - Calculus Single Variable

Question

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

Step-by-Step Solution

Answer

The first derivative is \( e^x(\sin(x) + \cos(x)) \), the second is \( 2e^x\cos(x) \), and the third is \( 2e^x(\cos(x) - \sin(x)) \).

Step 1: Identify the Function

The function given is \( f(x) = e^x \sin(x) \), which is a product of two functions. We will use the product rule to differentiate this function multiple times.

Step 2: Compute the First Derivative

The product rule states that if you have a function \( u(x)v(x) \), its derivative is \( u'(x)v(x) + u(x)v'(x) \). Here, let \( u(x) = e^x \) and \( v(x) = \sin(x) \). Then \( u'(x) = e^x \) and \( v'(x) = \cos(x) \). Thus, the first derivative is: \[ f'(x) = e^x \sin(x) + e^x \cos(x) = e^x (\sin(x) + \cos(x)) \]

Step 3: Differentiate Again to Get the Second Derivative

Using the product rule again, differentiate \( f'(x) = e^x (\sin(x) + \cos(x)) \). Let \( U(x) = e^x \) and \( V(x) = \sin(x) + \cos(x) \). Therefore, \( U'(x) = e^x \) and \( V'(x) = \cos(x) - \sin(x) \). Applying the product rule gives:\[f''(x) = e^x(\sin(x) + \cos(x)) + e^x(\cos(x) - \sin(x)) = e^x(2\cos(x)) = 2e^x\cos(x)\]

Step 4: Compute the Third Derivative

Now differentiate \( f''(x) = 2e^x\cos(x) \). Let \( u(x) = 2e^x \) and \( v(x) = \cos(x) \). Then, \( u'(x) = 2e^x \) and \( v'(x) = -\sin(x) \). Applying the product rule yields: \[f'''(x) = 2e^x \cos(x) + 2e^x(-\sin(x)) = 2e^x(\cos(x) - \sin(x))\]

Key Concepts

Product Rule

When dealing with the differentiation of products of two functions, the product rule is indispensable. This rule allows you to find the derivative of the product of two functions, denoted as \( u(x) \) and \( v(x) \). The rule states:

• If \( f(x) = u(x) \cdot v(x) \), then \( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

This means you differentiate the first function and multiply it by the second, then add it to the first function multiplied by the derivative of the second function. This approach ensures that every aspect of the product's rate of change is accounted for.
With this rule, you can systematically approach the differentiation of complex products like \( e^{x} \sin(x) \). Each differentiation follows the same pattern, making it a reliable tool for calculus problems.

First Derivative

To find the first derivative of a function like \( f(x) = e^{x} \sin(x) \), utilize the product rule:

• Set \( u(x) = e^{x} \) and \( v(x) = \sin(x) \).
• The derivatives are \( u'(x) = e^{x} \) and \( v'(x) = \cos(x) \).

Applying the product rule gives:\[f'(x) = e^{x} \sin(x) + e^{x} \cos(x) = e^{x}(\sin(x) + \cos(x))\]The first derivative, therefore, represents how the function \( e^{x} \sin(x) \) changes at any given point \( x \). It's like taking a snapshot of the function's rate of change, showing how the entire function behaves as \( x \) varies.
This step is fundamental because understanding the first derivative provides insights into the function’s increasing or decreasing nature.

Second Derivative

After determining the first derivative, \( f'(x) = e^{x}(\sin(x) + \cos(x)) \), we again apply the product rule for the second derivative, \( f''(x) \).

• Let \( U(x) = e^{x} \) and \( V(x) = \sin(x) + \cos(x) \).
• Then, \( U'(x) = e^{x} \) and \( V'(x) = \cos(x) - \sin(x) \).

By applying the product rule:\[f''(x) = e^{x}(\sin(x) + \cos(x)) + e^{x}(\cos(x) - \sin(x)) = e^{x}(2\cos(x)) = 2e^{x}\cos(x)\]The second derivative indicates the concavity of the function. It helps us understand how the rate of change itself changes, shedding light on the acceleration or the bending of the function graph.

Third Derivative

To find the third derivative, take the second derivative \( f''(x) = 2e^{x}\cos(x) \) and once again use the product rule:

• Set \( u(x) = 2e^{x} \) and \( v(x) = \cos(x) \).
• Then \( u'(x) = 2e^{x} \) and \( v'(x) = -\sin(x) \).

Applying the product rule gives:\[f'''(x) = 2e^{x} \cos(x) + 2e^{x}(-\sin(x)) = 2e^{x}(\cos(x) - \sin(x))\]The third derivative provides insight into the rate of change of the second derivative. It is often related to the jerk in physics - that is, the rate at which acceleration changes. Understanding higher-order derivatives helps in developing a more granular understanding of the function's behavior and its real-world applications.