Problem 5 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the derivative of each function. \(f(x)=(3 x+1)\left(x^{2}-2\right)\)

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = (3x + 1)(x^2 - 2)\) is \(f'(x) = 9x^2 + 2x - 6\).

Step 1: Identify the functions u and v

Let u be the first function and v be the second function: \( u(x) = (3x + 1) \) and \( v(x) = (x^2 - 2) \)

Step 2: Find the derivatives of u and v

Now, we need to find the derivatives of u and v. For u(x): \[ u'(x) = \frac{d}{dx} (3x + 1) = 3 \] For v(x): \[ v'(x) = \frac{d}{dx} (x^2 - 2) = 2x \]

Step 3: Apply the product rule to find the derivative of the given function

Using the product rule \( (uv)' = u'v + uv' \), we can calculate the derivative of f(x): \[ f'(x) = u'(x)v(x) + u(x)v'(x) = 3(x^2 - 2) + (3x + 1)(2x) \]

Step 4: Simplify the derivative

Now we just need to simplify the resulting expression: \[ f'(x) = 3x^2 - 6 + 6x^2 + 2x = (3 + 6)x^2 + 2x - 6 = 9x^2 + 2x - 6 \] Hence, the derivative of the function is \( f'(x) = 9x^2 + 2x - 6 \).

Key Concepts

Understanding Derivatives

Derivatives are fundamental in calculus. They help us understand how a function changes at any point. Imagine you are looking at a curve on a graph; the derivative tells you the slope, or steepness, of that curve at a specific point. This is like how steep a hill is if you were biking up or down.

When finding the derivative, you're essentially finding the rate at which one quantity changes in relation to another. In simpler terms, if you have a function \( f(x) \), the derivative \( f'(x) \) shows how \( f(x) \) changes as \( x \) changes.

Using derivatives, you can:

• Determine increasing or decreasing behavior of functions.
• Find max and min points of graphs, known as critical points.
• Solve real-world rate-change problems, like speed and acceleration.

Derivatives are everywhere in calculus and are essential for understanding motion, optimization, and much more.

Applying the Product Rule

The product rule is specifically used when you need to take the derivative of a function that is the product of two or more functions. Think of the product rule as a tool that lets you 'break down' complexity in calculus.

To apply the product rule, you identify two separate functions that multiply to form your original function. For example, if you have \( f(x) = u(x) \cdot v(x) \), you first find the derivatives of each function separately. You then use the product rule formula:

• \((uv)' = u'v + uv'\)

This formula is essential because it tells you how the combined functions change when both individual functions are changing at the same time.

In our example, the given function can be decomposed into:

• \( u(x) = 3x + 1 \)
• \( v(x) = x^2 - 2 \)

By using the product rule, you integrate these smaller parts to find the overall rate of change for the whole function. This approach is especially useful in physics and engineering, where systems are often comprised of interacting subsystems.

Simplification of Derivatives

Simplification is the last, but equally important, step in differentiating functions. When you apply the product rule, the resulting expression can often be complex, full of terms and multipliers.

Simplification involves combining like terms and reducing the expression to its most basic form. This makes further analysis much clearer and easier to handle.

For example, after using the product rule:

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

we simplify this by distributing and combining like terms:

• Start with \(3x^2 - 6 + 6x^2 + 2x\)
• Combine terms to get \((3 + 6)x^2 + 2x - 6 = 9x^2 + 2x - 6\)

Thus, the simplified derivative is \( f'(x) = 9x^2 + 2x - 6 \).

This approach not only makes it easier to interpret and use the derivative but also ensures accuracy in further calculations.