Problem 7 - Calculus for Biology and Medicine

Question

Differentiate the functions given with respect to the independent variable. $$ g(s)=5 s^{7}+2 s^{3}-5 s $$

Step-by-Step Solution

Answer

The derivative of the function is \( g'(s) = 35s^6 + 6s^2 - 5 \).

Step 1: Understand the Function

We have a polynomial function given by \( g(s) = 5s^7 + 2s^3 - 5s \). We need to differentiate this function with respect to \( s \).

Step 2: Differentiate Each Term

We differentiate each term of the polynomial separately. For a term of the form \( as^n \), the derivative is \( a \cdot n \cdot s^{n-1} \).

Step 3: Differentiate the First Term

The first term is \( 5s^7 \). Differentiating this term gives \( 5 \times 7 \times s^{7-1} = 35s^6 \).

Step 4: Differentiate the Second Term

The second term is \( 2s^3 \). Differentiating this term gives \( 2 \times 3 \times s^{3-1} = 6s^2 \).

Step 5: Differentiate the Third Term

The third term is \( -5s \). Differentiating this term gives \( -5 \times 1 \times s^{1-1} = -5 \).

Step 6: Combine the Derivatives

Combine the derivatives of each term to get the derivative of the whole function: \( g'(s) = 35s^6 + 6s^2 - 5 \).

Key Concepts

Understanding Polynomial Functions

Polynomial functions are a type of mathematical expression involving variables raised to whole number powers, with coefficients. Each segment of a polynomial function called a 'term' consists of a coefficient (a constant) multiplied by a variable raised to an exponent. For example, in the expression \( 5s^7 + 2s^3 - 5s \), \( 5s^7 \), \( 2s^3 \), and \( -5s \) are all individual terms.

These functions can take a variety of forms depending on the degree, which is the highest power of the variable present in the function. In our function \( g(s) = 5s^7 + 2s^3 - 5s \), the highest degree of \( s \) is 7, making this a seventh-degree polynomial. The general expression of a polynomial is \( a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \), where \( a_n \) are coefficients and \( n \) is a non-negative integer.

Understanding polynomial functions is essential for grasping differentiation as each term of the function can be differentiated independently owing to its simple structure.

Derivatives of Polynomial Terms

Differentiation is a process in calculus used to determine the rate at which a function is changing at any given point. When it comes to polynomial functions, each term can be differentiated individually using the rule: if you have a term in the form \( as^n \), then its derivative is \( n \cdot a \cdot s^{n-1} \). This is known as the power rule.

Let's take a closer look at each term in the expression \( 5s^7 + 2s^3 - 5s \) to see how differentiation works:

• The first term \( 5s^7 \) becomes \( 35s^6 \) after differentiation. You multiply the coefficient 5 by the exponent 7 to get 35, and decrease the exponent by 1.
• The second term \( 2s^3 \) becomes \( 6s^2 \). Similarly, you multiply the coefficient 2 by the exponent 3 to get 6, and the power is consequently reduced by one.
• The third term \( -5s \) becomes -5. Here, the exponent of the variable is 1, and when you reduce the exponent by 1, it becomes 0 (making \( s^0 \) equal to 1), so you just have the constant coefficient -5.

The beauty of the power rule is its simplicity and direct application, which makes it very practical for differentiating polynomial terms.

Differentiation in Calculus

Calculus is a significant branch of mathematics that focuses on limits, functions, derivatives, integrals, and infinite sequences and series. Differentiation, a core concept in calculus, is centered around finding the derivative of a function. The derivative signifies how a function's output changes concerning its input, essentially providing us the function's slope or steepness at any given point.

The process of differentiation can be applied to a wide array of functions, but it is generally straightforward for polynomial functions due to the simplicity of their terms. The derivative of a function like \( g(s) = 5s^7 + 2s^3 - 5s \) demonstrates how underlying patterns in a polynomial can reflect changes.

In practical applications, differentiation helps us understand the behavior of real-world phenomena, such as how fast a position changes over time (velocity) or how costs might increase with production levels. By mastering the differentiation of polynomials, one sets a solid foundation for tackling more complex challenges within calculus and in diverse practical situations.