Problem 5 - Calculus an Applied Approach

Question

Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$\begin{array}{ll}{\text { Function }} & {\text { Point }} \\\ {f(x)=\frac{1}{3}\left(2 x^{3}-4\right)} & {\left(0,-\frac{4}{3}\right)} \end{array}$$

Step-by-Step Solution

Answer

The value of the derivative of the function at the point \((0,-\frac{4}{3})\) is \(0\). The power rule was used to find the derivative.

Step 1: Differentiate the function f(x)

The function \(f(x)=\frac{1}{3}(2 x^{3}-4)\) can be simplified to \(f(x)=\frac{2}{3} x^{3}-\frac{4}{3}\). Then by using power rule, which states that the derivative of \(x^{n}\) is \(n*x^{n-1}\), the derivative \(f'(x)\) can be found: \(f'(x)=2x^{2}-0\). Therefore, the derivative \(f'(x)\) is \(2x^{2}\).

Step 2: Evaluate the derivative at the given point

To evaluate \(f'(x)\) at \(x=0\), substitute \(0\) into \(f'(x)\): \(f'(0)=2*(0)^{2}=0\). Thus, the value of the derivative of the function at the point \((0,-\frac{4}{3})\) is \(0\).

Key Concepts

Differentiation Rules

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any given point. Differentiation rules are the foundational methods that help us compute derivatives efficiently and accurately. When dealing with derivatives, it's crucial to choose the appropriate rule for a specific situation. These rules include:

• Power Rule: Used for functions where the variable is raised to a power.
• Product Rule: Applied when differentiating functions that are multiplied together.
• Quotient Rule: Necessary for functions that are divided.
• Chain Rule: Utilized for functions composed of other functions.

In solving a problem, recognizing the type of rule to apply is essential. For the function \(f(x)=\frac{2}{3} x^{3}-\frac{4}{3}\), we employed the Power Rule since the function involves a term with \(x\) raised to a power.

Power Rule

The Power Rule is one of the simplest and most frequently used differentiation rules in calculus. It provides a straightforward way to differentiate polynomial functions. The rule states that if you have a term \(x^n\), the derivative of this term with respect to \(x\) is \(n \cdot x^{n-1}\). Let's break this down with an example:

• Suppose you have \(f(x) = x^5\).
• Apply the Power Rule: Derivative \(f'(x) = 5x^{4}\).

In our original exercise, the function was \(f(x)=\frac{2}{3} x^{3}-\frac{4}{3}\). Applying the Power Rule:

• This would make the derivative of \(\frac{2}{3}x^3\) equal to \(3 \cdot \frac{2}{3}x^{3-1}\), simplifying to \(2x^2\).

By using this simple rule, we determined the rate of change of the function.

Derivative Evaluation

Once we have found the derivative of a function, the next step often involves evaluating this derivative at a specific point. This process involves simply substituting the given value of \(x\) into the derivative.In our exercise, after finding the derivative \(f'(x)=2x^2\), we needed to evaluate this at \(x=0\). To do this:

• Substitute \(x = 0\) in the derivative, which gives us \(f'(0) = 2(0)^2 = 0\).

Evaluating the derivative allows us to understand the behavior of the function at that specific point - in this case, confirming that the rate of change at \((0, -\frac{4}{3})\) is zero. This evaluation step is crucial in applications where determining the slope or rate at a particular point is necessary, such as finding tangent lines or verifying critical points.