Problem 20 - Calculus for Biology and Medicine

Question

Apply the product rule to find the tangent line, in slope-intercept form, of \(y=f(x)\) at the specified point. \(f(x)=\left(3 x^{3}-3\right)\left(2-2 x^{2}\right)\), at \(x=0\)

Step-by-Step Solution

Answer

The tangent line is \( y = -6 \).

Step 1: Identify the function components

First, let's identify the two functions being multiplied in the product rule. Let \[ u(x) = 3x^3 - 3 \] and \[ v(x) = 2 - 2x^2. \] We will differentiate each separately in the next steps.

Step 2: Differentiate u(x)

To find the derivative of \( u(x) = 3x^3 - 3 \), apply the power rule:\[ \frac{du}{dx} = 9x^2. \]

Step 3: Differentiate v(x)

Now, differentiate \( v(x) = 2 - 2x^2 \) using the power rule:\[ \frac{dv}{dx} = -4x. \]

Step 4: Apply the product rule

The product rule for differentiation states: \[ (uv)' = u'v + uv'. \]Applying this to the given function \( f(x) = u(x)v(x) \), we have:\[ f'(x) = (9x^2)(2 - 2x^2) + (3x^3 - 3)(-4x). \] Simplify it:\[ f'(x) = 18x^2 - 18x^4 - 12x^4 + 12x. \]

Step 5: Simplify the derivative

Combine like terms to simplify \( f'(x) \):\[ f'(x) = 18x^2 + 12x - 30x^4. \]

Step 6: Evaluate the derivative at x = 0

To find the slope at \( x = 0 \), substitute 0 into the derivative, \( f'(x) = 18x^2 + 12x - 30x^4 \):\[ f'(0) = 18(0)^2 + 12(0) - 30(0)^4 = 0. \]The slope of the tangent line at \( x = 0 \) is 0.

Step 7: Find the y-coordinate at x = 0

Substitute \( x = 0 \) into the original function \( f(x) \) to find the point:\[ f(0) = (3(0)^3 - 3)(2 - 2(0)^2) = -3 (2) = -6. \]The point on the curve where \( x = 0 \) is \( (0, -6) \).

Step 8: Write the equation of the tangent line

Using the point-slope form of a line equation, where the slope \( m = 0 \) and the point is \( (0, -6) \), the equation is:\[ y - (-6) = 0(x - 0). \]Simplifying gives:\[ y = -6. \] This is the equation of the tangent line in slope-intercept form \( y = mx + b \).

Key Concepts

Understanding the Product Rule

The product rule is a fundamental concept in calculus used to differentiate functions that are multiplied together. When you encounter a function expressed as a product of two simpler functions, such as \(f(x) = u(x) \cdot v(x)\), you need to differentiate each component and then combine the results. This is where the product rule comes in handy.

Mathematically, the product rule states that the derivative of the product of two functions, \(u(x)\) and \(v(x)\), is:\[ (uv)' = u'v + uv'. \]Here's a step-by-step breakdown:

• First, find the derivative of the first function \(u(x)\), denoted as \(u'\).
• Next, find the derivative of the second function \(v(x)\), denoted as \(v'\).
• Multiply \(u'\) by \(v\) and \(u\) by \(v'\).
• Add these two products together.

In the exercise you've been working on, breaking down the complex function into two parts made it easier to apply the product rule and find the derivative.

Exploring Tangent Lines

A tangent line is a straight line that touches a curve at exactly one point, without crossing it.

It's often used to approximate the behavior of the function near the point of tangency. The slope of the tangent line at a particular point gives us some insight into the rate of change of the function at that particular spot.

For the curve described by \(f(x)\), the process of finding a tangent line involves several key steps:

• First, calculate the derivative of the function. This gives us a new function that describes the slope of the tangent line for any given \(x\).-
• Next, substitute the \(x\)-value of the point where the tangent is needed into the derivative. This yields the slope of the tangent line at that specified point.
• The equation of the tangent line can then be found using this slope and the coordinates of the point of tangency.

Using Slope-Intercept Form

The slope-intercept form is a straightforward way of expressing the equation of a linear line. In mathematics, it's written as \(y = mx + b\), where:

• \(y\) is the dependent variable.
• \(m\) is the slope of the line, showing the rate of change.
• \(x\) is the independent variable.
• \(b\) is the y-intercept, which is the point where the line crosses the y-axis.

In the context of the exercise, after determining the slope from the derivative, we find the y-intercept by calculating the value of the function at the point of tangency. When you set the slope \(m\) and the intercept \(b\), writing the line equation in slope-intercept form becomes straightforward, as shown in the final steps where \(y = -6\) is derived.

Mastering Differentiation

Differentiation is the process of finding the derivative of a function, which tells us how the function's output changes in response to changes in the input. It's a central operation in calculus and essential for solving many problems involving rates of change and motion.

There are several rules and techniques, like the power rule, the product rule, and the chain rule, that can be used to differentiate different types of functions efficiently. Through differentiation, you can gain important information about the behavior of mathematical models without explicitly solving them.

• For simple polynomials, differentiating involves applying the power rule, which states that the derivative of \(x^n\) is \(nx^{n-1}\).
• When dealing with products of functions, the product rule becomes a powerful tool to find the derivative.
• Often, simplifying and combining terms after using these rules can help to clean up the result, making it easier to interpret.

In this exercise, differentiation helped us find both the slope of the tangent line and further the entire equation of the tangent, illustrating its application in finding derivatives of products.