Problem 35 - Calculus and its applications

Question

Differentiate. $$ f(x)=\left(x^{2}-2 x+2\right) e^{x} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = e^x(x^2 + 0) \). Correction: more carefully, \( f'(x) = e^x(x^2 + 2x) \).

Step 1: Identify the Functions for Product Rule

We are given the function \( f(x) = (x^2 - 2x + 2)e^x \). This is a product of two functions: \( u(x) = x^2 - 2x + 2 \) and \( v(x) = e^x \). We'll use the product rule for differentiation: \( (uv)' = u'v + uv' \).

Step 2: Differentiate \( u(x) \)

Find the derivative of \( u(x) = x^2 - 2x + 2 \). Using basic differentiation rules, the derivative \( u'(x) = 2x - 2 \).

Step 3: Differentiate \( v(x) \)

The function \( v(x) = e^x \) is differentiated using the exponential rule. So, \( v'(x) = e^x \).

Step 4: Apply the Product Rule

Using the product rule, substitute \( u'(x) \), \( v(x) \), and \( v'(x) \) found earlier: \( f'(x) = (2x - 2)e^x + (x^2 - 2x + 2)e^x \).

Step 5: Simplify the Expression

Factor \( e^x \) from the expression: \( f'(x) = e^x[(2x - 2) + (x^2 - 2x + 2)] \). Simplify the expression inside the brackets, \( (2x - 2) + (x^2 - 2x + 2) = x^2 + 0x = x^2 \), so \( f'(x) = e^x(x^2) \). However, note we missed combining like terms correctly in the simplification, so it should be \( (2x - 2) + (x^2 - 2x + 2) = x^2 + 0 = x^2 \), but the careful recombination shows no further term reduction beyond step calculation previously.The actual final expression results once recombined correctly as:\( f'(x) = e^x(x^2 + 2x - 2) \).

Key Concepts

Product Rule

In calculus, differentiation is the process of finding the rate at which a function is changing at any given point. The product rule is a key part of this process, especially useful when dealing with the derivative of the product of two functions. When you have a function that is the product of two functions, say \( f(x) = u(x) \cdot v(x) \), where \( u(x) \) and \( v(x) \) are both differentiable, you cannot simply take the derivative of each function individually and multiply them. Instead, the product rule comes to the rescue. Each function contributes in a specific way: \((uv)' = u'v + uv'\), meaning you differentiate the first function \( u(x) \), leaving the second function \( v(x) \) unchanged, then add it to the product of the first function unchanged and the differentiation of the second. In simpler terms, the derivation goes like this:

• First, find the derivative of \( u(x) \), and multiply by \( v(x) \).
• Then, find the derivative of \( v(x) \), and multiply by \( u(x) \).
• Lastly, add both expressions together.

This will give you the derivative of \( f(x) = u(x) \cdot v(x) \). This method ensures you're accounting for every way the functions influence each other's rate of change.

Derivative of Exponential Functions

Exponential functions are a very special type of mathematical function, typically expressed as \( e^x \), where \( e \) is a constant approximately equal to 2.71828. A key property of exponential functions is that their differentiation results in the same function. When you differentiate \( v(x) = e^x \), the result is surprisingly simple and elegant: \( v'(x) = e^x \). This property is unique to exponential functions and makes them much easier to handle in calculus compared to many other types of functions. Another form that often comes across is \( a^x \, (a > 0) \). For differentiation, you'd use the rule \( (a^x)' = a^x \ln(a) \), which extends the exponential rule beyond the natural base \( e \).When dealing with exponential functions in calculus:

• The derivative of the natural exponent, \( e^x \), is \( e^x \) itself.
• If the exponential function includes a coefficient like \( ce^x \), then \( (ce^x)' = ce^x \), where \( c \) is a constant.

This characteristic is what makes exponential functions so powerful and widely applicable in real-world scenarios, from compound interest calculations to population growth models.

Basic Differentiation Rules

Differentiation builds upon basic rules that simplify finding derivatives of more complex expressions. Understanding these basic rules simplifies the process of working through calculus problems. Here are the essential rules of differentiation to keep in mind:

• **Power Rule:** If \( f(x) = x^n \), then the derivative \( f'(x) = n x^{n-1} \). This rule applies when you have a variable raised to a power.
• **Constant Rule:** If \( f(x) = c \), where \( c \) is constant, then \( f'(x) = 0 \). Constant functions have zero rate of change.
• **Constant Multiple Rule:** If \( f(x) = c \cdot u(x) \), then \( f'(x) = c \cdot u'(x) \). This rule helps when a function is multiplied by a constant.
• **Sum/Difference Rule:** If \( f(x) = u(x) \pm v(x) \), then \( f'(x) = u'(x) \pm v'(x) \). To differentiate a sum or difference of functions, you differentiate each function separately and then add or subtract their derivatives respectively.

In the original problem, the power rule was used to find the derivative of \( u(x) = x^2 - 2x + 2 \). Applying the power and sum/difference rules, the derivative is obtained as \( u'(x) = 2x - 2 \). Familiarity with these rules not only helps with finding derivatives directly but also simplifies tackling more complex functions by breaking them down into more manageable parts.