Problem 37 - Brief Applied Calculus

Question

Use the Generalized Power Rule to find the derivative of each function. $$ f(x)=(2 x+1)^{3}(2 x-1)^{4} $$

Step-by-Step Solution

Answer

The derivative is \((2x+1)^2(2x-1)^3 (28x+2)\).

Step 1: Identify the Structure

To find the derivative of the function \( f(x)=(2x+1)^3(2x-1)^4 \), we first note that this is a product of two functions, \( u(x)=(2x+1)^3 \) and \( v(x)=(2x-1)^4 \). Therefore, we'll need to use the product rule in conjunction with the generalized power rule.

Step 2: Apply the Product Rule

The product rule states that if \( f(x) = u(x)v(x) \), then the derivative \( f'(x) = u'(x)v(x) + u(x)v'(x) \). So, we'll compute \( u'(x) \) and \( v'(x) \) separately.

Step 3: Differentiate \( u(x) \) Using the Generalized Power Rule

For \( u(x) = (2x+1)^3 \), use the generalized power rule: \( \frac{d}{dx}[(g(x))^n] = n(g(x))^{n-1}g'(x) \). Here, \( n = 3 \) and \( g(x) = 2x+1 \). Thus, \( u'(x) = 3(2x+1)^2 \cdot 2 = 6(2x+1)^2 \), where \( g'(x) = 2 \).

Step 4: Differentiate \( v(x) \) Using the Generalized Power Rule

For \( v(x) = (2x-1)^4 \), using the same rule, \( n = 4 \) and \( g(x) = 2x-1 \). So, \( v'(x) = 4(2x-1)^3 \cdot 2 = 8(2x-1)^3 \), where \( g'(x) = 2 \).

Step 5: Substitute into the Product Rule Formula

Substitute \( u'(x) \), \( v'(x) \), \( u(x) \), and \( v(x) \) back into the product rule formula: \[ f'(x) = 6(2x+1)^2(2x-1)^4 + (2x+1)^3 \times 8(2x-1)^3 \]

Step 6: Simplify the Expression

Both terms of \( f'(x) \) share common factors. To simplify, factor out the greatest common factor: \( (2x+1)^2(2x-1)^3 \). Thus, \[ f'(x) = (2x+1)^2(2x-1)^3 \left[6(2x-1) + 8(2x+1)\right] \]. Simplify the expression inside the brackets: \[ 6(2x-1) + 8(2x+1) = 12x - 6 + 16x + 8 = 28x + 2 \]. So, the simplified derivative is \[ f'(x) = (2x+1)^2(2x-1)^3 (28x+2) \].

Key Concepts

Understanding Derivatives

The concept of a derivative is foundational in calculus. A derivative measures how a function changes as its input changes. This can be thought of as the slope of the function's graph at any given point. For instance, if you have a function representing the distance an object travels over time, the derivative would give you the object's speed.Mathematically, the derivative is found using a process called differentiation. The derivative of a function is denoted by symbols such as \( f'(x) \) or \( \frac{df}{dx} \). It tells us the rate of change of the function with respect to its variable, typically \( x \).

• If the derivative is positive, the function is increasing.
• If the derivative is negative, the function is decreasing.
• If the derivative is zero, the function may have a local maximum, minimum, or a point of inflection.

To compute derivatives efficiently, rules such as the power rule are used, particularly helpful for functions in polynomial form.

Exploring the Product Rule

When a function is composed of two or more multiplied functions, like \( f(x) = u(x) \cdot v(x) \), we need to use the product rule to find its derivative. The product rule is a key tool in calculus that allows us to differentiate products of functions.The product rule states:\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]

• \( u'(x) \) is the derivative of the first function, \( u(x) \).
• \( v'(x) \) is the derivative of the second function, \( v(x) \).

Substituting these into the product rule helps us find the derivative of the composite function. This rule effectively breaks down complex differentiation problems into simpler parts, making the process manageable.

Mastering the Differentiation Process

Differentiation is the process of finding the derivative of a function. This involves a series of steps and the application of various rules to simplify and solve.For power functions, the Generalized Power Rule is particularly helpful. This rule states that the derivative of a function \( (g(x))^n \) is:\[ \frac{d}{dx}[(g(x))^n] = n(g(x))^{n-1}g'(x) \]

• \( n \) is the exponent.
• \( g(x) \) is the inner function.
• \( g'(x) \) is the derivative of the inner function.

In the presented exercise, the function \( f(x)=(2x+1)^3(2x-1)^4 \) required breaking down using both the product rule and applying the generalized power rule to each component. By systematically differentiating each part and combining them, we find the derivative \( f'(x) \). This structured approach ensures accuracy and clarity when solving complex calculus problems.