Problem 39 - College Algebra and Calculus: An Applied Approach

Question

find \(f^{\prime}(x)\). $$ f(x)=x\left(x^{2}+1\right) $$

Step-by-Step Solution

Answer

The derivative \(f^{\prime}(x)\) of the function \(f(x)=x(x^{2}+1)\) is \(3x^{2}+1\).

Step 1: Identify the Functions for the Product Rule

We have \(f(x)=x(x^{2}+1)\) which is equivalent to \(f(x)=u(x)v(x)\) where \(u(x)=x\) and \(v(x)=x^{2}+1\). These two functions will be used when applying the product rule.

Step 2: Differentiate Function \(u(x)\)

Let's find the derivative of \(u(x)\). \(u^{\prime}(x)=\frac{d}{dx}(x)=1\).

Step 3: Differentiate Function \(v(x)\)

Now we find the derivative of \(v(x)\). \(v^{\prime}(x)=\frac{d}{dx}(x^{2}+1)=2x\).

Step 4: Apply the Product Rule

The product rule is given by \((uv)^{\prime}=u^{\prime}v+uv^{\prime}\). Applying the product rule, we find \(f^{\prime}(x)=u^{\prime}(x)v(x) + u(x)v^{\prime}(x) = (1)(x^{2}+1) + x(2x) = x^{2}+1+2x^{2}=3x^{2}+1\).

Key Concepts

Derivative of Functions

Understanding the derivative of a function is a fundamental concept in calculus. At its core, the derivative represents the rate at which a function's value is changing at any given point. For instance, if we consider a function that denotes the distance traveled by a car over time, the derivative of this function would give us the car's speed at any given moment.

Mathematically, if we have a function like that in our exercise, namely, \( f(x) = x(x^2+1) \), finding the derivative—denoted as \( f'(x) \)—is about finding that instantaneous rate of change. When the function is composed of simple power expressions, as in our example, we can generally apply standard differentiation rules like the power rule. However, when functions are multiplied together, we need to use specific techniques—like the product rule—to accurately calculate the derivative.

Applying the Product Rule

When a function \( f(x) \) is the product of two sub-functions, such as \( u(x) \) and \( v(x) \), the product rule is imperative for finding the derivative. This rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. This can be expressed as \( (uv)' = u'v + uv' \).

In the context of the exercise, where we have \( f(x) = x(x^2+1) \), the product rule guides us through the process seamlessly. After differentiating \( u(x) = x \) and \( v(x) = x^2+1 \) individually, we apply the product rule to combine these derivatives into the solution for \( f'(x) \). This systematic approach ensures that students can tackle even more complex functions involving products.

Differentiation Techniques

Differentiation techniques are methods used to find the derivative of a function. Mastery of these techniques is vital for students to navigate through calculus efficiently. In addition to the product rule, there are several other key techniques:

• The Power Rule for functions like \( x^n \) which simply become \( nx^{n-1} \).
• The Quotient Rule which is used when a function is a quotient of two other functions.
• The Chain Rule which is applied when a function is the composition of two or more functions.
• Higher Order Derivatives which involve taking the derivative of a derivative, and so on.

Applying these techniques judiciously, as per the structure of the function at hand, allows for accurate and efficient calculation of derivatives. For instance, in our current exercise, identifying that the product rule was the appropriate technique was crucial in arriving at the correct solution for \( f'(x) \).