Problem 12 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Differentiate the following. (a) \(f(x)=\frac{x^{2} e^{x}}{3}\) (b) \(f(x)=\frac{5 x^{2}}{3 e^{2}}\) (c) \(f(x)=\frac{1}{x e^{5 x}}\)

Step-by-Step Solution

Answer

The derivatives of the provided functions are (a) \( f'(x) = \frac{x^{2} e^{x}}{3} \) , (b) \( f'(x) = \frac{10x}{9e^{2}} \), and (c) \( f'(x) = -\frac{e^{5x} + 5xe^{5x}}{(x e^{5x})^{2}} \).

Step 1: Differentiating function (a)

The given function is \( f(x)=\frac{x^{2} e^{x}}{3} \). Here, \( u = x^{2} \) and \( v = e^{x} \). Calculate the derivatives of \( u \) and \( v \) to get \( u' = 2x \) and \( v' = e^{x} \). Apply the quotient rule to find \( f'(x) = \frac{3 * x^{2}*e^{x} - x^{2}*2xe^{x}}{3^{2}} \). After simplifying, we get \( f'(x) = \frac{x^{2} e^{x}}{3} \).

Step 2: Differentiating function (b)

The given function is \( f(x)=\frac{5 x^{2}}{3 e^{2}} \). Here, \( u = 5x^{2} \) and \( v = 3e^{2} \). Since \( v \) is a constant, the derivative is zero. So, simply differentiate \( u \) to get \( u' = 10x \). Hence, \( f'(x) = \frac{v * u' - u * 0}{v^{2}} \). After simplifying, we get \( f'(x) = \frac{10x}{9e^{2}} \).

Step 3: Differentiating function (c)

The given function is \( f(x)=\frac{1}{x e^{5 x}} \). Here, \( u = 1 \) and \( v = xe^{5x} \). Derivating \( u \) we get \( u' = 0 \) and differentiating \( v \) using the product rule we get \( v' = e^{5x} + 5xe^{5x} \). So, \( f'(x) = \frac{v*u' - u*v'}{v^{2}} \). After simplifying, we get \( f'(x) = -\frac{e^{5x} + 5xe^{5x}}{(x e^{5x})^{2}} \).

Key Concepts

Quotient Rule

The quotient rule is essential when working with the differentiation of a function that is a division of two other functions. Let's say we have a function in the form of \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \). To differentiate this, the quotient rule states:

• \( \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \)

This rule allows us to calculate the derivative of a ratio of two functions effectively. Remember that \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \), respectively.
In the provided solution for functions \( (a) \) and \( (c) \), the quotient rule was used. In function \( (a) \), the derivative of the product \( x^2 e^x \) sits in the numerator, alongside a constant in the denominator. Correctly applying this rule allows us to handle more complex derivative problems.
By following these steps, each component of the function can be differentiated effectively, providing a correct solution for both complex and simple rational functions.

Product Rule

The product rule is used when differentiating a product of two differentiable functions. If \( u(x) \) and \( v(x) \) are two functions, then their product's derivative can be found using the equation:

• \( (uv)' = u'v + uv' \)

This means you take the derivative of the first function and multiply it by the second function, and vice versa.
In practical terms, think of it as breaking up the differentiation process into two parts. First, differentiate the first part of the product while keeping the second part constant and then flip it: keep the first part constant and differentiate the second part.
In our example, the product rule comes into play in Step 3, while finding the derivative of the function \( xe^{5x} \) in function \( (c) \). Therefore, understanding the product rule helps you manage equations where the product of functions are involved, reducing complexity during the differentiation process.

Derivative of Exponential Functions

Exponential functions, such as \( e^x \), have unique properties that make their differentiation straightforward. When differentiating \( e^{ax} \), where \( a \) is a constant, the derivative is given by:

• \( \frac{d}{dx}(e^{ax}) = ae^{ax} \)

You retain the exponential form, which simplifies a lot of calculus problems.
In the given problems, differentiating \( e^x \) and \( e^{5x} \) were key steps. You simply multiply the entire exponential by its exponent's derivative.
By applying this rule, you can find derivatives quickly. It's especially useful in physical sciences and engineering where exponential growth and decay models frequently arise. Thus, mastering this can make tackling exponential functions less daunting.

Simplifying Derivatives

Simplifying derivatives is a crucial final step in differentiation, making the result easier to understand and use in further calculations. Once the derivative has been found, simplification often involves:

• Combining like terms
• Reducing fractions
• Factoring common factors

Effective simplification ensures that the derivative is presented in its most concise form, which aids in understanding and application.
In the examples given, each derivative was simplified to more compact forms to make the result tidier and more useful. For instance, in function \( (c) \), factoring and canceling terms reduced the complexity considerably.
Practicing simplification not only strengthens the understanding of algebra but also enhances the clarity of a solution in calculus. It's beneficial in translating complex results into usable mathematical information, crucial for application in real-life scenarios.