Problem 39 - Calculus an Applied Approach

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the functions

We first identify the two functions in the given function \(f(x) = x(x^{2}+1)\). Here, the first function is \(u(x)=x\) and the second function is \(v(x)=x^{2} + 1\).

Step 2: Compute the derivatives

Next, we find the derivatives of both identified functions. The derivative of the function \(u(x) = x\) is \(u^{\prime}(x) = 1\) and the derivative of the function \(v(x) = x^{2} + 1\) is \(v^{\prime}(x) = 2x\).

Step 3: Apply the product rule

Now, we will apply the product rule. According to the product rule, the derivative of a product of two functions is given by \(f^{\prime}(x) = u^{\prime}(x) * v(x) + u(x) * v^{\prime}(x)\). Substituting the values of \(u^{\prime}(x)\), \(v^{\prime}(x)\), \(u(x)\), and \(v(x)\), we get: \(f^{\prime}(x) = 1*(x^{2} + 1) + x*2x\). Simplify this to get the final result.

Step 4: Simplify the function

Simplify the function to get: \(f^{\prime}(x) = x^{2} + 1 + 2x^{2}\). Combine like terms to get: \(f^{\prime}(x) = 3x^{2} + 1\)

Key Concepts

Product Rule

The Product Rule is a critical tool in calculus used to find the derivative of a product of two functions. In our exercise, we deal with the function \( f(x) = x(x^{2}+1) \). Here, it's evident that we need to calculate the derivative by considering the separate functions \( u(x) = x \) and \( v(x) = x^2 + 1 \). The rule states that if you have two differentiable functions \( u(x) \) and \( v(x) \), then the derivative of their product is given by:

• \( f^{\prime}(x) = u^{\prime}(x) \cdot v(x) + u(x) \cdot v^{\prime}(x) \)

Applying this in our context, we first need to find the derivatives of the individual functions. Then, we substitute these derivatives into the formula. This way, we effectively break down a potentially complex problem into simpler, more manageable parts by handling the differentiation one function at a time.

Derivative of Polynomial Functions

Calculating the derivative of polynomial functions is a fundamental skill in calculus. Each term in a polynomial is typically in the form \( ax^n \), where \( a \) is a constant and \( n \) is the exponent. The process involves applying the power rule, which states that for a function \( x^{n} \), the derivative is \( nx^{n-1} \). This rule allows us to quickly and effectively find the derivative of any polynomial term by multiplying the coefficient by the exponent and reducing the exponent by one.
In our problem, we find the derivative of \( v(x) = x^2 + 1 \) as an example. The derivative \( v^{\prime}(x) \) becomes \( 2x \) because the term \( x^2 \) transforms into \( 2x \) using the power rule, and the derivative of a constant \(1\) is zero. For \( u(x) = x \), the derivative \( u^{\prime}(x) \) is simply \( 1 \), because the term \( x \) reduces to \( x^{0} \), which is 1. This straightforward approach makes it easy to handle even complex polynomial expressions by tackling each term individually.

Simplifying Expressions

Once you've found the derivative using the product rule, the next essential step is simplifying the expression. Simplification is crucial because it makes the expression much easier to interpret and use in further calculations or real-world applications. In our example, we begin with \( f^{\prime}(x) = 1(x^{2} + 1) + x(2x) \).

• First, distribute the terms: \( f^{\prime}(x) = x^{2} + 1 + 2x^2 \).
• Next, combine like terms by adding up the coefficients of the \( x^{2} \) terms: \( x^{2} + 2x^{2} \) simplifies to \( 3x^{2} \).
• So the final simplified expression becomes \( f^{\prime}(x) = 3x^2 + 1 \).

Simplifying expressions effectively means looking for common factors, combining like terms, and making the expression as concise as possible. By taking these steps, you ensure that your solution is not only correct but also neat and comprehensible.