Problem 3 - 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. $$\begin{array}{ll}{\text { Function }} & {\text { Point }} \\\ {f(x)=x^{2}\left(3 x^{3}-1\right)} & {(1,2)} \end{array}$$

Step-by-Step Solution

Answer

The value of the derivative of the function \(f(x) = x^2(3x^3 - 1)\) at the point (1,2) is 13.

Step 1: Identify the Functions

Given function is \(f(x) = x^2(3x^3 - 1)\). We can consider \(u(x) = x^2\) and \(v(x) = 3x^3 - 1\). Hence, \(f(x) = u(x) * v(x)\).

Step 2: Apply the Product Rule

The derivative of the function \(f'(x) = u'(x) * v(x) + u(x) * v'(x)\). The derivative of \(u(x) = x^2\) is \(u'(x) = 2x\) and the derivative of \(v(x) = 3x^3 - 1\) is \(v'(x) = 9x^2\). \[f'(x) = 2x * (3x^3 - 1) + x^2 * 9x^2 = 6x^4 - 2x + 9x^4 = 15x^4 - 2x\]

Step 3: Evaluate the derivative at the point

The value of the derivative at the point (1,2) is obtained by substituting \(x = 1\) into the derivative \(f'(x) = 15x^4 - 2x\). Therefore, \(f'(1) = 15(1)^4 - 2(1) = 15 - 2 = 13\).

Key Concepts

Product Rule

When dealing with functions that are multiplied together, the product rule is essential for finding derivatives. This rule applies to functions in the form of \(f(x) = u(x) \cdot v(x)\). It's a handy tool when you're working with the derivative of a product of two functions. The product rule states that if you have two differentiated functions \(u(x)\) and \(v(x)\), their derivatives follow this pattern:

• \(f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\)

This means you first differentiate \(u(x)\) and multiply by the original \(v(x)\), then you take the original \(u(x)\) and multiply by the derivative of \(v(x)\). Sum these two results to get the derivative of \(f(x)\). This rule simplifies the process of finding the derivative of two intertwined functions, allowing us to deal with more complex expressions easily.

Differentiation Rules

Differentiation rules are guidelines that help us find the derivative of different types of functions. These rules are foundational in calculus and include the power rule, product rule, quotient rule, and chain rule, among others. Each rule has unique applications, making it easier to handle derivatives in various situations.
For our specific case, the product rule is used because we're dealing with the multiplication of two functions, \(x^2\) and \(3x^3 - 1\). Yet, in finding the derivatives of \(u(x) = x^2\) and \(v(x) = 3x^3 - 1\), we utilize the power rule. The power rule states that the derivative of \(x^n\) is \(n \cdot x^{n-1}\), which helps us quickly differentiate terms involving powers of \(x\).
Keep in mind that these rules not only provide a step-by-step framework but also enhance our understanding of how functions change, making differentiation a powerful tool for analyzing real-world scenarios.

Calculating Derivatives

Calculating derivatives is a systematic process of applying differentiation rules to break down a function into its rate of change. In our example, the function \(f(x) = x^2(3x^3 - 1)\) requires us to follow specific steps to calculate \(f'(x)\):

• First, identify \(u(x) = x^2\) and \(v(x) = 3x^3 - 1\).
• Differentiate both: \(u'(x) = 2x\) and \(v'(x) = 9x^2\).
• Apply the product rule: \(f'(x) = 2x \cdot (3x^3 - 1) + x^2 \cdot 9x^2\).
• Simplify to get \(f'(x) = 15x^4 - 2x\).

Following these steps ensures accuracy and clarity in finding the derivative. Each step builds on the previous one, leading to the final simplified derivative.

Evaluating Derivatives at a Point

Evaluating derivatives at a specific point provides insight into the behavior of a function at that exact location. This process involves substituting the point of interest into the derivative to get an exact value. In our exercise, we want to evaluate \(f'(x)\) at the point \((1, 2)\).
To do this, we substitute \(x = 1\) into the derived function \(f'(x) = 15x^4 - 2x\). After substituting, simplify to find:

• \(f'(1) = 15 \cdot (1)^4 - 2 \cdot 1 = 15 - 2 = 13\)

This gives a precise slope of the tangent line to the curve at \(x = 1\). This result indicates the rate of change of the original function \(f(x)\) at this particular point, helping us understand how the function behaves instantaneously at \(x = 1\).