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. $$h(s)=(2 s-1)(s+4)$$

Short answer

The derivative is \( 4s + 7 \), verified by both the product rule and expansion.

Step-by-step solution

Apply the Product Rule

The Product Rule for differentiation states that if you have two functions, say \( u(x) \) and \( v(x) \), then the derivative \( (uv)' \) is given by \( u'v + uv' \). Apply this to the given function \( h(s) = (2s-1)(s+4) \). Let \( u = 2s-1 \) and \( v = s+4 \). Differentiating each, we have \( u' = 2 \) and \( v' = 1 \). Thus the derivative is:\[ h'(s) = (2)(s+4) + (2s-1)(1) = 2s + 8 + 2s - 1 = 4s + 7 \].

Algebraic Manipulation of the Function

Expand the function \( h(s) = (2s-1)(s+4) \) to simplify it first. Start by distributing the terms:\[ h(s) = 2s(s+4) - 1(s+4) = 2s^2 + 8s - s - 4 = 2s^2 + 7s - 4 \]. This gives us a simpler polynomial form of \( h(s) \).

Differentiate the Simplified Function

Now differentiate the simplified polynomial \( h(s) = 2s^2 + 7s - 4 \). Use the power rule for each term:- The derivative of \( 2s^2 \) is \( 4s \).- The derivative of \( 7s \) is \( 7 \).- The derivative of a constant \(-4\) is \( 0 \).Thus, the derivative is:\[ h'(s) = 4s + 7 \].

Verify Equivalence of the Results

Compare the derivatives obtained from Steps 1 and 3. Both are \( h'(s) = 4s + 7 \). This shows that both methods yield the same result, confirming the correctness of the calculations.

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus used to determine the rate at which a function is changing at any point. In essence, it computes the slope of the tangent line to the curve of a function.
When differentiating a product of two functions, we often use the Product Rule, but differentiation can also be streamlined by simplifying expressions first. This helps to uncover easier forms of the function that can be differentiated using basic rules like the power rule.
Consistently applying differentiation enables comparisons and verifications of results, ensuring that the calculations align regardless of the method being used.

Algebraic Manipulation

Algebraic manipulation involves transforming expressions into a different but equivalent form. This often makes them easier to work with. In our exercise, we started with the function \( h(s) = (2s-1)(s+4) \). Instead of directly using complex rules, we expanded this expression:

• Distribute each term in the first bracket through the terms in the second bracket.
• This results in \( h(s) = 2s^2 + 8s - s - 4 \).
• Simplify to \( h(s) = 2s^2 + 7s - 4 \).

Algebraic manipulation transforms a product into a polynomial, making future steps in differentiation simpler and more efficient.

Power Rule

The power rule is used to differentiate functions of the form \( ax^n \). To apply it, multiply the coefficient \( a \) by the exponent \( n \), and reduce the exponent by one.
When we applied the power rule to the simplified function \( 2s^2 + 7s - 4 \), we considered each term separately.

• For \( 2s^2 \), differentiate to get \( 4s \).
• For \( 7s \) (which is \( 7s^1 \)), differentiate to get \( 7 \).
• The constant \( -4 \) becomes \( 0 \) as constants don't change and, therefore, have no rate of change.

By applying the power rule, we're able to find the derivative of each term systematically, arriving at \( h'(s) = 4s + 7 \).

Equivalent Expressions

Equivalent expressions are different representations of the same quantity and are crucial for verifying mathematical results.
In this exercise, we verified that the derivative using the Product Rule, \( h'(s) = 4s + 7 \), matched the derivative from differentiating a simplified expression derived through algebraic manipulation.
This equivalence confirms that different approaches can coexist robustly.

• Using multiple methods gives confidence in the result.
• Both the applied Product Rule and algebraic manipulation led to the same derivative, ensuring the integrity of the process.

Thus, confirming results in mathematics strengthens understanding and supports learning objectives.