Problem 16 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the second derivative of the function. $$ h(s)=s^{3}\left(s^{2}-2 s+1\right) $$

Step-by-Step Solution

Answer

The second derivative of the function is \( h''(s) = 20s^{3} - 24s^{2} + 6s \).

Step 1: Apply Product Rule

First applying the product rule on \( s^{3}(s^{2} - 2s + 1) \) yields \( h'(s) = 3s^{2}(s^{2} - 2s + 1) + s^{3}(2s - 2) = 3s^{4}- 6s^{3} + 3s^{2} + 2s^{4} - 2s^{3} \). Simplifying gives \( h'(s) = 5s^{4} - 8s^{3} + 3s^{2} \).

Step 2: Apply Power Rule

Next, applying the power rule to find the second derivative gives \( h''(s) = 20s^{3} - 24s^{2} + 6s \).

Key Concepts

Understanding the Product Rule

The product rule is a crucial concept in calculus when it comes to differentiating products of two functions. Imagine you have two functions, say \( u(x) \) and \( v(x) \), and they're multiplied together. To differentiate their product, we need to employ the product rule. This rule tells us that the derivative of \( u(x) \cdot v(x) \) is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). Simply put, differentiate the first and multiply by the second, then add the product of the first and the derivative of the second. This method ensures you accurately represent the rate of change of the combined function. In our specific exercise, we applied the product rule to \( s^3(s^2 - 2s + 1) \). Here's how it unfolded: - Differentiate \( s^3 \) to get \( 3s^2 \), and multiply it by \( (s^2 - 2s + 1) \). - Then, keep \( s^3 \) as is and differentiate \( (s^2 - 2s + 1) \), which results in \( 2s - 2 \).- Combine these results to get the new expression after simplification. This part of the process helps set up the function for further differentiation.

Mastering the Power Rule

The power rule is one of the simplest rules in calculus for finding derivatives, but it's very powerful. It's used when you need to differentiate functions in the form of \( x^n \), where \( n \) is any real number. The rule states that the derivative of \( x^n \) is \( nx^{n-1} \). This means you bring down the power as a coefficient and then subtract one from the original power.In the exercise, after using the product rule, we found a new polynomial: \( 5s^4 - 8s^3 + 3s^2 \). To find the second derivative, we applied the power rule:- For \( 5s^4 \), the derivative is \( 20s^3 \).- For \( -8s^3 \), it becomes \( -24s^2 \).- Finally, for \( 3s^2 \), the derivative is \( 6s \).Using these, we combined them to get the second derivative \( 20s^3 - 24s^2 + 6s \). The power rule simplifies the process of finding higher order derivatives, offering a straightforward way to handle polynomial terms.

Exploring Polynomial Differentiation

Polynomial differentiation is a fundamental aspect of calculus used frequently because many functions are polynomials. A polynomial is simply a sum of terms, each consisting of a variable raised to a power multiplied by a coefficient. When dealing with polynomials, each term is differentiated separately using rules like the power rule. In our problem, the function \( h(s) = s^3(s^2 - 2s + 1) \) was initially a product of polynomials. After applying the product rule and finding an expanded polynomial form, differentiating them individually using the power rule became straightforward. - This involved handling each term of the polynomial \( 5s^4 - 8s^3 + 3s^2 \) independently, allowing for the simple sum of derivatives.Polynomial differentiation enables us to assess the behavior of dynamic systems, model economic trends, and even predict scientific outcomes. Since you break down each term separately to differentiate, it becomes a mechanical yet insightful task to uncover the behavior of complex functions.