Problem 8 - Calculus for Biology and Medicine

Question

Differentiate the functions given in Problems with respect to the independent variable. $$ g(s)=3-4 s^{2}-4 s^{3} $$

Step-by-Step Solution

Answer

The derivative of \( g(s) \) is \( g'(s) = -8s - 12s^2 \).

Step 1: Recognize the Problem Type

We are asked to differentiate the function \( g(s) = 3 - 4s^2 - 4s^3 \) with respect to the variable \( s \). This is a basic differentiation problem where we apply standard differentiation rules.

Step 2: Apply Differentiation Formula

For any function \( g(s) = a_n s^n + a_{n-1} s^{n-1} + \, ... \, + a_1 s + a_0 \), the derivative is given by \( g'(s) = n a_n s^{n-1} + (n-1) a_{n-1} s^{n-2} + \, ... \, + a_1 \). This is known as the power rule of differentiation, \( \frac{d}{ds}s^n = ns^{n-1} \).

Step 3: Differentiate Each Term of the Function

Differentiate each term of the function \( g(s) = 3 - 4s^2 - 4s^3 \):- The derivative of \( 3 \) with respect to \( s \) is \( 0 \) because the derivative of a constant is zero.- The derivative of \( -4s^2 \) is \( -8s \) applying the power rule, as \( -4 \times 2 \times s^{2-1} = -8s \).- The derivative of \( -4s^3 \) is \( -12s^2 \) by similarly applying the power rule, \( -4 \times 3 \times s^{3-1} = -12s^2 \).

Step 4: Combine the Derivatives

Combine all the derivatives to get the overall derivative:\[g'(s) = 0 - 8s - 12s^2 = -8s - 12s^2.\]

Step 5: Write the Final Answer

The derivative of the function \( g(s) = 3 - 4s^2 - 4s^3 \) with respect to \( s \) is given by:\[g'(s) = -8s - 12s^2.\]

Key Concepts

Power Rule

The power rule is a fundamental concept in differentiation. It provides a straightforward way to find the derivative of a power function. This rule is particularly useful when dealing with polynomials. The power rule states that if you have a term of the form \( s^n \), where \( n \) is any real number, its derivative is \( ns^{n-1} \). This means you bring down the exponent as a coefficient and then reduce the exponent by one.

For example, using the power rule, the derivative of \( s^3 \) is \( 3s^2 \). Each term in a polynomial is handled the same way. This rule significantly simplifies the process of differentiation, especially when dealing with long expressions.

The power rule can be applied to each term of a polynomial separately and then those derivatives can be combined for the entire expression. This makes it a go-to technique for many differentiation problems.

Basic Differentiation

Basic differentiation involves using a set of rules to determine the rate of change of a function. In differentiation, each term of a polynomial is differentiated separately. The derivative of a constant is always zero because constants do not change. Differentiation is essentially finding how the function changes as the variable changes.

Some fundamental rules of differentiation include:

• The derivative of a constant function is zero.
• The derivative of a sum of functions is the sum of their derivatives.
• Using the power rule, as discussed, for each term.

These simple rules lay the foundation for more complex calculus operations, supporting the calculation of slopes, rates of change, and many other concepts central to calculus.

Derivative of a Polynomial

Polynomials are a type of function that can be differentiated using basic rules of calculus. Derivatives measure how a function changes as its input changes, and for polynomials, this process is systematic and efficient.

To find the derivative of a polynomial, apply the power rule to each term:

• If a polynomial has the form \( a_n s^n + a_{n-1} s^{n-1} + \, ... \, + a_1 s + a_0 \), apply the power rule to each term.
• For example: Given \( g(s) = 3 - 4s^2 - 4s^3 \)
• The derivative of \( 3 \) is \( 0 \), since the derivative of a constant is zero.


• The derivative of \( -4s^2 \) is \( -8s \).


• The derivative of \( -4s^3 \) is \( -12s^2 \).

After finding the derivatives of individual terms, linear terms are simply added together to produce the final expression, which in this case yields \( g'(s) = -8s - 12s^2 \).

This method offers a clear pathway to understanding and solving polynomial differentiation problems.