Problem 1 - Applied Calculus

Question

If \(f(x)=(2 x+1)(3 x-2)\), find \(f^{\prime}(x)\) two ways: by using the product rule and by multiplying out. Do you get the same result?

Step-by-Step Solution

Answer

Yes, both methods give \( f'(x) = 12x - 1 \).

Step 1: Identify f(x)

The function is given as \( f(x) = (2x + 1)(3x - 2) \). To find the derivative \( f'(x) \), we will use two methods: the Product Rule and expanding the polynomial first.

Step 2: Use the Product Rule

The Product Rule states that the derivative of a product \( u(x) \cdot v(x) \) is given by \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). Here, let \( u(x) = 2x + 1 \) and \( v(x) = 3x - 2 \).

Step 3: Find Derivatives u'(x) and v'(x)

Calculate the derivatives: \( u'(x) = 2 \) and \( v'(x) = 3 \).

Step 4: Apply the Product Rule

Apply the Product Rule: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = 2 \cdot (3x - 2) + (2x + 1) \cdot 3 \].

Step 5: Simplify the Expression Using the Product Rule

Simplify \( f'(x) \): \[ f'(x) = 2(3x - 2) + 3(2x + 1) = 6x - 4 + 6x + 3 = 12x - 1 \].

Step 6: Expand f(x)

Now, expand \( f(x) = (2x + 1)(3x - 2) \) to simplify it before differentiating. Perform the multiplication: \[ (2x + 1)(3x - 2) = 6x^2 - 4x + 3x - 2 \].

Step 7: Simplify the Expanded Expression

Combine like terms: \[ 6x^2 - 4x + 3x - 2 = 6x^2 - x - 2 \].

Step 8: Differentiate the Expanded Expression

Differentiate \( 6x^2 - x - 2 \): \[ f'(x) = \frac{d}{dx}(6x^2) - \frac{d}{dx}(x) - \frac{d}{dx}(2) = 12x - 1 \].

Step 9: Verify the Result

Both methods give the same derivative \( f'(x) = 12x - 1 \). This confirmation shows our calculations are consistent and correct.

Key Concepts

Derivative

A derivative is a fundamental concept in calculus. It measures how a function changes as its input changes. Think of it as the function's rate of change or its slope at a particular point. For example, if you have the function \( f(x) = (2x + 1)(3x - 2) \), you're interested in finding \( f'(x) \), which is the derivative of \( f(x) \). This tells us how \( f(x) \) changes for small changes in \( x \).

• If the derivative is positive, the function is increasing.
• If it's negative, the function is decreasing.
• A zero derivative can indicate a local minimum, maximum, or saddle point.

Understanding derivatives is crucial for comprehending other calculus topics like integrals and differential equations.

Product Rule

The product rule is a useful technique for finding the derivative of a product of two functions. When you have two functions, \( u(x) \) and \( v(x) \), and you want to find the derivative of their product \( u(x) \cdot v(x) \), you use the product rule.

Formula of the Product Rule

To apply it, remember the formula: \[f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\]This formula tells us we need to find:

• The derivative of the first function \( u'(x) \)
• The second function \( v(x) \)
• The first function \( u(x) \)
• The derivative of the second function \( v'(x) \)

Then, simply plug these values into the formula and simplify to find the derivative of the product.
The product rule is especially helpful when both functions are more complicated polynomial expressions or trigonometric functions.

Polynomial Expansion

Polynomial expansion involves multiplying expressions to simplify them, helping make differentiating easier. In our example, we start with \( f(x) = (2x + 1)(3x - 2) \). By expanding, we distribute each term from one polynomial to every term of the other, like this:

Steps in Polynomial Expansion

• Multiply \( 2x \) by \( 3x \) and \( -2 \)
• Multiply \(+1\) by \( 3x \) and \( -2 \)

This results in the expression: \( 6x^2 - 4x + 3x - 2 \).

Simplifying Polynomials

Combining like terms simplifies it to \( 6x^2 - x - 2 \). With the simplified polynomial, differentiating becomes straightforward, as polynomials are among the easiest functions to differentiate. Each term is differentiated individually, yielding \( f'(x) = 12x - 1 \).
Understanding polynomial expansion is vital when dealing with products of sums or more complex polynomial expressions.