Question

(a) Use the Product Rule to differentiate the function. (b) Manipulate the function algebraically and differentiate without the Product Rule. (c) Show that the answers from (a) and (b) are equivalent. $$f(x)=x\left(x^{2}+3 x\right)$$

Short answer

Both methods give the derivative: \( 3x^2 + 6x \).

Step-by-step solution

Recognize the Function's Structure

The given function is a product of two functions: \( f(x) = x imes (x^2 + 3x) \).Identify that we will apply the product rule for differentiation in part (a) and expand the expression for part (b).

Apply the Product Rule

The Product Rule states that if you have two functions \( u(x) \) and \( v(x) \), then their product's derivative is \((uv)' = u'v + uv' \).Here, let \( u(x) = x \) and \( v(x) = x^2 + 3x \).Differentiate each:\( u'(x) = 1 \),\( v'(x) = 2x + 3 \).Therefore, \((uv)' = u'(x)v(x) + u(x)v'(x) = 1(x^2 + 3x) + x(2x + 3) \).Simplify to obtain the derivative: \( = x^2 + 3x + 2x^2 + 3x = 3x^2 + 6x \).

Algebraically Manipulate the Function

Expand the original product before differentiating:\( f(x) = x(x^2 + 3x) = x^3 + 3x^2 \).Now, differentiate this expanded function: \(f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(3x^2) \).

Differentiate Without Product Rule

Differentiate each term separately:From \( x^3 \), \(\frac{d}{dx}(x^3) = 3x^2 \).From \( 3x^2 \), \(\frac{d}{dx}(3x^2) = 6x \).Adding these derivatives, we get: \( 3x^2 + 6x \).

Compare the Results

The derivatives obtained in steps (2) and (4) are both \( 3x^2 + 6x \).This confirms that the two methods produce equivalent results.

Key concepts

Product Rule

When differentiating a product of two functions, we use the Product Rule. This rule is essential when dealing with functions that are multiplied together, as it helps us find the derivative of the entire expression. Here's how it works:

• If you have two functions, say \( u(x) \) and \( v(x) \), the derivative of their product is given by: \[(uv)' = u'v + uv'\]
• In this exercise, we identify \( u(x) = x \) and \( v(x) = x^2 + 3x \). So, to apply the Product Rule, we need to find the derivative of each function separately.

First, the derivative of \( u(x) = x \) is \( u'(x) = 1 \).
The derivative of \( v(x) = x^2 + 3x \) is \( v'(x) = 2x + 3 \).By substituting these into the formula, we get:\[u'(x) = 1 \, (x^2 + 3x) + x \, (2x + 3)\]This simplifies to \( 3x^2 + 6x \). This demonstrates how the Product Rule simplifies handling more complex functions.

Function Expansion

To differentiate without using the Product Rule, one can first simplify the function by expanding it. Function expansion involves multiplying out factors so the expression no longer looks like a product. This helps make differentiating more straightforward.For the function we are working with, \( f(x) = x(x^2 + 3x) \), expanding it turns it into a polynomial:

• First, distribute \( x \) across the terms inside the parentheses:\[f(x) = x \cdot x^2 + x \cdot 3x = x^3 + 3x^2 \]
• This new expression is now a simple polynomial, easy to handle with standard differentiation techniques.

Next, differentiate each term individually:

• The derivative of \( x^3 \) is \( 3x^2 \).
• The derivative of \( 3x^2 \) is \( 6x \).

So, combining these results gives us the same derivative: \( 3x^2 + 6x \). This alternative method of expansion simplifies calculations, especially when the Product Rule might not be immediately apparent.

Equivalent Derivatives

Equivalence in derivatives means that no matter which method you choose for differentiation, the results should be the same. This concept assures the reliability and consistency of calculus techniques. In our function, both the Product Rule and Function Expansion are used to find the derivative. Despite the different approaches, comparing the two results is key:

• Using the Product Rule: The derivative was found to be \( 3x^2 + 6x \).
• Using Function Expansion and simple differentiation: The derivative was also \( 3x^2 + 6x \).

This shows that both methods give identical outcomes.
This equivalence is critical in calculus, confirming that mathematical transformations don’t affect the ultimate derivative of the function.Students should feel confident that whether they choose to apply the Product Rule or simplify the function first, the result should be the same. This assurance allows flexibility in problem-solving and enhances comprehension of fundamental calculus principles.