Question

Finding and Evaluating a Derivative In Exercises \(13-18,\) find \(f^{\prime}(x)\) and \(f^{\prime}(c) .\) $$ \text{Function} \quad \text {Value of} \quad c $$ $$ f(x)=\left(x^{3}+4 x\right)\left(3 x^{2}+2 x-5\right) \quad c=0 $$

Short answer

The derivative \(f'(x) = 6x^{4} + 14x^{3} + 14x^{2} + 8x - 20\), and evaluated at \(c=0\) is \(f'(0) = -20\).

Step-by-step solution

Identify the Functions

In our case, we have two functions which are multiplied together. We can identify them as \(u(x) = x^{3} + 4x\) and \(v(x) = 3x^{2} + 2x - 5\).

Differentiate the Functions

We find the derivatives of \(u(x)\) and \(v(x)\). The derivative of the function \(u(x) = x^{3} + 4x\) is \(u'(x) = 3x^{2} + 4\). The derivative of the function \(v(x) = 3x^{2} + 2x - 5\) is \(v'(x) = 6x + 2\).

Apply the Product Rule

We now apply the rule \( \frac{d}{dx} [u(x)v(x)] = u(x)v'(x) + u'(x)v(x)\). Here \(u(x) = x^{3} + 4x, v(x) = 3x^{2} + 2x - 5, u'(x) = 3x^{2} + 4, v'(x) = 6x + 2\). So the derivative \(f'(x) = u(x)v'(x) + u'(x)v(x) = (x^{3} + 4x)(6x + 2) + (3x^{2} + 4)(3x^{2} + 2x - 5)\) which simplifies to \(f'(x) = 6x^{4} + 14x^{3} + 14x^{2} + 8x - 20\).

Evaluate the Derivative at the Point

Lastly, to find \(f'(c)\) for a specific \(c = 0\), we substitute \(x = 0\) into \(f'(x) = 6x^{4} + 14x^{3} + 14x^{2} + 8x - 20\) to get \(f'(0) = -20\).

Key concepts

Derivative

A derivative is a fundamental concept in calculus. It represents the rate at which a function is changing at any given point. In essence, a derivative can be thought of as a way to describe how a particular function behaves when you make small changes to the input value. This is incredibly useful because it allows us to understand the behavior of functions beyond single data points.

For example, if you envision a curve representing distance over time, the slope of that curve at any specific time would give you the speed at that moment. That's essentially the derivative. To find a derivative, we often use various rules and methods, such as the product rule, which is employed when dealing with products of two functions. This exercise involves finding the derivative of a function composed of two polynomial functions, which we'll explore further in the next sections.

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. They have a general form:

• Constant function: f(x) = c
• Linear function: f(x) = ax + b
• Quadratic function: f(x) = ax^2 + bx + c
• Cubic function: f(x) = ax^3 + bx^2 + cx + d

Each term in a polynomial function consists of a coefficient and a variable raised to a power, known as a **degree**. For instance, in the function \(x^3 + 4x\), the highest degree is 3, making it a cubic polynomial.

A polynomial's behavior is partly dictated by its degree, which determines the number of roots and the shape of its graph. Identifying the degree helps when differentiating, as it tells us how many times we will need to apply the derivative. Each time we differentiate, we decrease the power of x by 1, simplifying the polynomial function further. The combination of two polynomials, such as \((x^3 + 4x)\) and \((3x^2 + 2x - 5)\) in the example, is a bit more complex but follows similar principles with added rules like the product rule.

Differentiation

Differentiation is the process of finding the derivative of a function. It involves applying rules such as the power rule, product rule, quotient rule, and chain rule. In this exercise, we utilize the product rule, which provides a mechanism for finding the derivative of a product of two functions.

The product rule states that if you have two functions, \(u(x)\) and \(v(x)\), their derivative can be found using the formula:\[\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + u'(x)v(x)\]Applying this to our functions \(u(x) = x^3 + 4x\) and \(v(x) = 3x^2 + 2x - 5\), we calculate their derivatives individually and then insert them into the formula. This allows us to determine the overall derivative of their product, \(f(x)\).

Discovering \(f'(x) = 6x^4 + 14x^3 + 14x^2 + 8x - 20\) shows how each component contributes to the shape and direction of the curve. Lastly, by substituting specific values of \(x\) into \(f'(x)\), we can evaluate the derivative at those points, offering valuable insights into the function's behavior at particular moments, for instance, \(f'(0) = -20\).