Problem 58 - Calculus Single Variable

Question

Let \(f_{n}(x)=x^{n} e^{x}\) for every positive integer \(n\). Find \(f_{n}^{\prime}\) in terms of \(f_{n}\) and \(f_{n-1}\).

Step-by-Step Solution

Answer

\(f_n^{\prime}(x) = f_n(x) + nf_{n-1}(x)\).

Step 1: Understand the Problem

We need to differentiate the function \(f_n(x) = x^n e^x\). Our goal is to express the derivative \(f_n^{\prime}(x)\) in terms of \(f_n(x)\) and \(f_{n-1}(x)\).

Step 2: Apply the Product Rule

We recognize that \(f_n(x)\) can be seen as a product of two functions: \(u(x) = x^n\) and \(v(x) = e^x\). According to the product rule, the derivative \((uv)^{\prime} = u^{\prime}v + uv^{\prime}\).

Step 3: Differentiate Components

First, find the derivative of \(u(x) = x^n\) which is \(u^{\prime}(x) = nx^{n-1}\). Next, find the derivative of \(v(x) = e^x\) which is \(v^{\prime}(x) = e^x\).

Step 4: Apply the Product Rule

Using the product rule: \[ f_n^{\prime}(x) = x^n (e^x)^{\prime} + (x^n)^{\prime} e^x = x^n e^x + nx^{n-1} e^x. \]

Step 5: Express in terms of \(f_n\) and \(f_{n-1}\)

Now, since \(f_n(x) = x^n e^x\) and \(f_{n-1}(x) = x^{n-1} e^x\), rewrite the previous expression as: \[ f_n^{\prime}(x) = f_n(x) + nf_{n-1}(x) \].

Step 6: Finalize the Solution

Thus, \(f_n^{\prime}(x)\) is expressed in terms of \(f_n(x)\) and \(f_{n-1}(x)\) as \(f_n^{\prime}(x) = f_n(x) + nf_{n-1}(x)\).

Key Concepts

Product Rule

The product rule is a crucial technique in calculus for differentiating expressions where two functions are multiplied together. When dealing with a function that is a product of two separate functions, say \( u(x) \) and \( v(x) \), the product rule allows us to find the derivative of the product without directly differentiating the product itself. It states that the derivative of such a product is:

• \((uv)^{\prime} = u^{\prime}v + uv^{\prime}\)

This means, to find the derivative of the entire product, you:

• First, differentiate \( u(x) \) to get \( u^{\prime}(x) \).
• Then, differentiate \( v(x) \) to get \( v^{\prime}(x) \).
• Finally, plug these derivatives into the formula: multiply \( u^{\prime}(x) \) by \( v(x) \), and \( u(x) \) by \( v^{\prime}(x) \), then add both results.

This is a fundamental rule needed across many differentiation problems, especially when dealing with products of polynomials and exponential functions.

Derivative of Exponential Functions

Exponential functions, typically of the form \( e^x \), have unique and useful properties when it comes to differentiation. One of the most fascinating characteristics of exponential functions is that they are their own derivatives. This means:

• If \( v(x) = e^x \), then \( v^{\prime}(x) = e^x \).

What does this mean practically? When you differentiate an exponential function, it remains unchanged. In more complex functions where an exponential is multiplied by another polynomial or variable, such as \( x^n e^x \), you'll have to apply the product rule as well. Regardless of how complex the attached polynomial may be, maintaining the exponential part becomes crucially simple with this derivative rule. This makes exponential functions much easier to work with in calculus problem-solving compared to other functions, which might change form upon differentiation.

Calculus Problem-Solving

When approaching any calculus problem, a structured method of problem-solving is essential. Here is a simple approach to help.

• Understand the problem: Begin by analyzing the given function. In our example, \( f_n(x) = x^n e^x \), you need to determine what needs to be differentiated.
• Identify applicable rules: Notice the function involves a product and an exponential. It’s clear the product rule and the principle of the derivative of exponential functions will be required.
• Derive systematically: Start by differentiating each part of the function as required by the product rule. Then, using what you know about derivative properties, assemble the complete derivative.
• Express final solution: Using known relationships or function definitions to simplify or express the solution in a desired form, like \( f_n^{\prime}(x) = f_n(x) + nf_{n-1}(x) \).

This logical step-by-step procedure simplifies complex calculus tasks into manageable pieces and helps in gaining deeper insights into how differential calculus works. Adopting effective problem-solving strategies allows for significant improvement in confidently tackling calculus problems.