Problem 35 - Brief Applied Calculus

Question

Use the Generalized Power Rule to find the derivative of each function. $$ f(x)=3 x^{2}(2 x+1)^{5} $$

Step-by-Step Solution

Answer

The derivative is \(f'(x) = (2x + 1)^4 (30x^2 + 12x)\).

Step 1: Recognize the Functions

Notice that the function given is a product of two functions: \(u(x) = 3x^2\) and \(v(x) = (2x + 1)^5\). We need to find the derivative \(f'(x)\) of this product using the Product Rule and the Generalized Power Rule.

Step 2: Apply the Product Rule

The derivative of a product \(u(x)v(x)\) is \(u'(x)v(x) + u(x)v'(x)\). First, identify the derivatives \(u'(x)\) and \(v'(x)\).

Step 3: Differentiate \(u(x) = 3x^2\)

Using the Power Rule, the derivative \(u'(x) = 6x\).

Step 4: Differentiate \(v(x) = (2x + 1)^5\)

Using the Generalized Power Rule, which states \((g(x)^n)' = n \cdot g(x)^{n-1} \cdot g'(x)\), first find \(g(x) = 2x + 1\) and its derivative \(g'(x) = 2\). This gives \(v'(x) = 5(2x + 1)^4 \, \cdot 2 = 10(2x + 1)^4\).

Step 5: Combine Using the Product Rule

Substitute \(u'(x)\), \(v(x)\), \(u(x)\), and \(v'(x)\) into the Product Rule equation: \(f'(x) = u'(x)v(x) + u(x)v'(x) = 6x(2x + 1)^5 + 3x^2 \cdot 10(2x + 1)^4\).

Step 6: Simplify the Expression

Factor out the common term \((2x + 1)^4\) from both parts of the expression: \(f'(x) = (2x + 1)^4 [6x(2x + 1) + 30x^2]\), which simplifies to \(f'(x) = (2x + 1)^4 (12x + 30x^2)\).

Step 7: Final Simplification

Combine the terms inside the square bracket: \(f'(x) = (2x + 1)^4 (30x^2 + 12x)\). This can be further factored if necessary, but is correct as is.

Key Concepts

Product Rule

The Product Rule is an essential part of calculus, particularly when finding the derivative of a function that is the product of two other functions. If you have a function expressed as a product, such as \( f(x) = u(x) \cdot v(x) \), the Product Rule tells you how to find its derivative. The formula is as follows: the derivative of the function \( f(x) \) is \( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \). This formula comes from the need to take into account both the rate of change of \( u(x) \) while \( v(x) \) remains constant and the change in \( v(x) \) while \( u(x) \) is constant.

• Identify both functions that are multiplied: \( u(x) \) and \( v(x) \).
• Find the derivatives \( u'(x) \) and \( v'(x) \).
• Substitute into the formula \( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

This rule breaks down the potentially complex function into simpler parts, making it easier to handle.

Derivative of a Function

Understanding what a derivative means is crucial in calculus. A derivative reflects the rate at which a function is changing at any given point, which can be thought of as its instantaneous rate of change or the slope of the tangent line to the curve at that point. To find a derivative, one must understand the rules of differentiation, including the Power Rule and the Product Rule, as they guide us in differentiating various types of functions. The process of finding a derivative typically involves:

• Recognizing the type and structure of the original function.
• Applying appropriate differentiation rules—like the Power Rule for polynomials or the Product Rule for products of functions.
• Performing algebraic manipulation to simplify the resulting expression.

When correctly applied, these principles allow us to find the derivative of even complex functions efficiently.

Power Rule

The Power Rule is one of the simplest and most commonly used rules in differentiation, and it's used to find the derivative of functions of the form \( f(x) = x^n \). According to the Power Rule, the derivative of \( x^n \) is \( n \cdot x^{n-1} \). This rule makes it easy to differentiate polynomials and is an important tool in calculus for working through derivatives quickly. However, in some scenarios involving functions raised to a power, such as \( f(x) = (g(x))^n \), you'd use the Generalized Power Rule. Here, differentiating such a function involves more steps:

• Identify the inner function \( g(x) \) and calculate its derivative \( g'(x) \).
• Use the Generalized Power Rule: \( (g(x)^n)' = n \cdot g(x)^{n-1} \cdot g'(x) \).

By following these steps, you can find the derivative efficiently. Both the basic and generalized versions of the Power Rule are critical tools for understanding and performing differentiation.