Problem 8 - 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. $$\begin{array}{ll}{\text { Function }} & {\text { Point }} \\\ {g(x)=\left(x^{2}-2 x+1\right)\left(x^{3}-1\right)} & {(1,0)} \end{array}$$

Step-by-Step Solution

Answer

The value of the derivative of the function at the given point (1,0) is 0.

Step 1: Identify the Two Functions

In our case, the two functions that are being multiplied together are \(f(x) = x^{2}-2x+1\) and \(g(x) = x^{3}-1\).

Step 2: Apply the Product Rule

The product rule states that the derivative of two functions multiplied together is the first function times the derivative of the second plus the second function times the derivative of the first. So \((f \cdot g)' = f'\cdot g + f \cdot g'\). Using this rule, the derivative of our function is \((g(x))' = f'(x) \cdot g(x) + f(x) \cdot g'(x)\). Now we need to find the derivatives of \(f(x)\) and \(g(x)\). \[f'(x) = 2x-2, g'(x) = 3x^{2}\] So \[(g(x))' = (2x-2) \cdot (x^{3}-1) + (x^{2}-2x+1) \cdot 3x^{2}\]

Step 3: Simplify the Derivative

Perform the multiplication and addition operations: \((g(x))' = 2x^{4} - 2x - 2x^{3} + 2 + 3x^{4} - 6x^{3} + 3x^{2} = 5x^{4} - 8x^{3} + 3x^{2} + 2 - 2x\). This is the derivative of the function \(g(x)\).

Step 4: Evaluate the Derivative at the Given Point

Now, we need to evaluate the derivative at the point (1,0). So \((g'(1)) = 5(1)^{4} - 8(1)^{3} + 3(1)^{2} + 2 - 2(1) = 5 - 8 + 3 + 2 - 2 = 0\).

Key Concepts

product_rule

In calculus, the product rule is a crucial tool when you're dealing with functions that are multiplied together. It helps you find the derivative of a product of functions. Suppose you have two differentiable functions, denoted as \( f(x) \) and \( g(x) \). The product rule tells us that the derivative of these functions, expressed as \((f \cdot g)'\), is given by:

• \( f'(x) \cdot g(x) + f(x) \cdot g'(x) \)

The key idea here is to differentiate each function, one at a time, while keeping the other function as is. This allows you to find the slope or rate of change of the product of these two functions.
It might seem a bit confusing at first, but the good news is that it's mainly a matter of identifying the appropriate parts in the formula and substituting the derivatives accordingly.

differentiation

Differentiation is a fundamental concept in calculus, which involves finding the derivative of a function. The derivative is essentially the rate of change or the slope of the function at any given point.
In the context of our exercise, differentiation helped us to work out the derivatives of both \( f(x) = x^2 - 2x + 1 \) and \( g(x) = x^3 - 1 \).
The derivative of \( f(x) \), denoted as \( f'(x) \), is calculated by applying basic differentiation rules to each term:

• \( (x^2)' = 2x \)
• \((-2x)' = -2 \)
• \(1' = 0 \)

Resulting in \( f'(x) = 2x - 2 \).
Similarly, for \( g(x) \), we use the power rule for each term:

• \( (x^3)' = 3x^2 \)
• \((-1)' = 0 \)

Giving us \( g'(x) = 3x^2 \). You apply these calculated derivatives to substitute into the product rule later.

evaluate_at_a_point

Once you have the derivative of a function using the product rule and differentiation, the next step is often to evaluate it at a specific point. This means substituting the given value of \( x \) into your derivative to find out the slope or rate of change at that particular \( x \) value.
In our exercise, we were asked to evaluate the derivative of the function at the point \((1,0)\). This simply requires substituting \( x = 1 \) into the derivative we found:

• The derivative was \( 5x^4 - 8x^3 + 3x^2 + 2 - 2x \).
• Substitute: \( 5(1)^4 - 8(1)^3 + 3(1)^2 + 2 - 2(1) = 0 \).

This calculation shows that at \( x = 1 \), the derivative, which represents the rate of change of the given function, is 0. This indicates that the slope of the function is flat, or zero, at \( x = 1 \). It's a helpful process especially when you're exploring how a function behaves at particular points.