Question

Derivatives a. Use limits to find the derivative function \(f^{\prime}\) for the following functions \(f\) b. Evaluate \(f^{\prime}(a)\) for the given values of \(a\) $$f(s)=4 s^{3}+3 s ; a=-3,-1$$

Short answer

Question: Find the derivative function \(f'(s)\) for the given function \(f(s) = 4s^3 + 3s\), and evaluate the derivative for \(a = -3\) and \(a = -1\). Solution: The derivative function is \(f'(s) = 12s^2 + 12s + 3\), and the evaluated derivative values are: - For \(a = -3\): \(f^{\prime}(-3) = 75\) - For \(a = -1\): \(f^{\prime}(-1) = 3\)

Step-by-step solution

Recall the definition of a derivative

The definition of a derivative is given by: $$f^{\prime}(s)=\lim _{h \rightarrow 0} \frac{f(s+h)-f(s)}{h}$$

Apply the limit definition to find \(f'(s)\)

Plug the given function \(f(s)=4s^3+3s\) into the limit definition: $$f^{\prime}(s)=\lim _{h \rightarrow 0} \frac{(4(s+h)^3+3(s+h))-(4s^3+3s)}{h}$$ Now, expand and simplify the expression: $$f^{\prime}(s)=\lim _{h \rightarrow 0} \frac{4(s^3+3s^2h+3sh^2+h^3)+3(s+h)-4s^3-3s}{h}$$ $$f^{\prime}(s)=\lim _{h \rightarrow 0} \frac{12s^2h+12sh^2+4h^3+3h}{h}$$ Now, factor out h: $$f^{\prime}(s)=\lim _{h \rightarrow 0} h(12s^2+12s+4h+3)$$

Evaluate the limit

We now have: $$f^{\prime}(s)=h\lim _{h \rightarrow 0}(12s^2+12s+4h+3)$$ As \(h\) approaches 0, the term \(4h\) becomes 0. So, we are left with: $$f^{\prime}(s)=12s^2+12s+3$$ We have found the derivative function \(f'(s)\).

Evaluate \(f'(a)\) for the given values of \(a\)

To find \(f'(a)\) for \(a = -3\) and \(a = -1\), we plug these values into our expression for \(f'(s)\): For \(a=-3\): $$f^{\prime}(-3)=12(-3)^2+12(-3)+3$$ $$f^{\prime}(-3)=108-36+3$$ $$f^{\prime}(-3)=75$$ For \(a=-1\): $$f^{\prime}(-1)=12(-1)^2+12(-1)+3$$ $$f^{\prime}(-1)=12-12+3$$ $$f^{\prime}(-1)=3$$

Final Answer

The derivative function \(f'(s) = 12s^2 + 12s + 3\), and the values of the derivative are: $$f^{\prime}(-3) = 75$$ $$f^{\prime}(-1) = 3$$

Key concepts

Limit Definition of Derivative

The concept of a derivative is a fundamental part of calculus. It represents the rate at which a function is changing at any given point. Imagine a car traveling down a highway; the derivative tells you the car's speed at any moment.

To find the derivative of a function at a particular point using the limit definition, we use the equation: \[f^{\prime}(s)=\lim _{h \rightarrow 0} \frac{f(s+h)-f(s)}{h}\]. This formula measures the average rate of change of the function as we consider two points getting infinitely close.

As an example, for a function \(f(s)\), we need to find \(f(s+h)\) and \(f(s)\) and substitute them into the formula. By taking the limit as \(h\) approaches 0, we effectively find the instantaneous rate of change, or the derivative, at \(s\). It's like zooming in so close to a curve that it almost looks like a straight line.

Polynomial Functions

Polynomial functions are mathematical expressions consisting of variables raised to whole number exponents and multiplied by coefficients.

This can be something like \(4s^3 + 3s\) which is our example function here.

Elements of a polynomial include:

• Terms: Each part of the summation separated by plus or minus signs (e.g., \(4s^3\) and \(3s\)).
• Coefficients: The numbers in front of the variables (e.g., 4 and 3).
• Exponents: The powers the variables are raised to (e.g., 3 and 1).

Polynomials are a vital part of calculus because their derivatives are easy to compute and often result in another polynomial, allowing for simplifications and recognitions of patterns in behavior as you calculate derivatives.

Evaluating Derivatives

Once we have the general derivative formula from the limit definition, the next step is to evaluate it at specific points. This means we substitute given values into our derivative function.

The derivative provides the slope of the tangent line to the function at a specific point. Using our example, \(f'(s) = 12s^2 + 12s + 3\), we find the rate of change at two points, \(a = -3\) and \(a = -1\).

For \(a = -3\), substitute into the derivative formula: \[f^{\prime}(-3) = 12(-3)^2 + 12(-3) + 3 = 108 - 36 + 3 = 75\]. For \(a = -1\), substitute into the formula: \[f^{\prime}(-1) = 12(-1)^2 + 12(-1) + 3 = 12 - 12 + 3 = 3\].

• The result \(75\) tells us how steeply the function is climbing when \(s = -3\).
• The result \(3\) gives us the slope when \(s = -1\).

Evaluating the derivatives gives us a set of answers that indicate how fast or slow the function's value is changing at these specific inputs.