Problem 11 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the derivative of each function. \(f(x)=\left(5 x^{2}+1\right)(2 \sqrt{x}-1)\)

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = (5x^2 + 1)(2\sqrt{x} - 1)\) is: \(f'(x) = 20x\sqrt{x} - 10x + \frac{5x^2}{\sqrt{x}} + \frac{1}{\sqrt{x}}\)

Step 1: Find the derivative of u(x)

First, let's find the derivative of u(x) with respect to x: \(u(x) = 5x^2 + 1\) To find \(u'(x)\), differentiate u(x) with respect to x: \(u'(x) = 10x\)

Step 2: Find the derivative of v(x)

Next, let's find the derivative of v(x) with respect to x: \(v(x) = 2\sqrt{x} - 1\) To find \(v'(x)\), differentiate v(x) with respect to x: \(v'(x) = \frac{d}{dx}(2\sqrt{x} - 1) = 2 \frac{d}{dx}(\sqrt{x}) - \frac{d}{dx}(1)\) Now, we know that \(\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}\), so we substitute this into the equation to find \(v'(x)\): \(v'(x) = 2 \cdot \frac{1}{2\sqrt{x}} - 0\) Simplify v'(x): \(v'(x) = \frac{1}{\sqrt{x}}\)

Step 3: Apply the product rule

Now, we can apply the product rule to find the derivative of f(x): \((u \cdot v)' = u' \cdot v + u \cdot v'\) Substitute the derivatives of u(x) and v(x) that we found earlier: \(f'(x) = (10x)(2\sqrt{x} - 1) + (5x^2 + 1)(\frac{1}{\sqrt{x}})\) Now we'll simplify the equation: \(f'(x) = 20x\sqrt{x} - 10x + \frac{5x^2}{\sqrt{x}} + \frac{1}{\sqrt{x}}\)

Step 4: Final answer

The derivative of the given function is: \(f'(x) = 20x\sqrt{x} - 10x + \frac{5x^2}{\sqrt{x}} + \frac{1}{\sqrt{x}}\)

Key Concepts

Product Rule

When dealing with derivatives in calculus, especially when two functions are multiplied together, the product rule is essential. Imagine you have two functions, say \( u(x) \) and \( v(x) \), and you need to find the derivative of their product \( f(x) = u(x) \cdot v(x) \). The product rule provides a method to do so without having to expand the functions. The rule states:

• \((u \cdot v)' = u' \cdot v + u \cdot v'\)

This formula tells us that we differentiate each function separately, multiply each derivative by the other original function, and then sum these products.
This approach is incredibly efficient, especially when the functions become complex. With our example, we identified \(u(x) = 5x^2 + 1\) and \(v(x) = 2\sqrt{x} - 1\). By applying the product rule, we combined their derivatives \(u'(x)\) and \(v'(x)\) in a straightforward manner, leading us to the final derivative result. This streamlined process emphasizes why knowing the product rule is valuable for tackling diverse calculus problems.

Differentiation

Differentiation is one of the core operations in calculus. It involves finding the derivative of a function, which represents the rate at which the function's value changes with respect to changes in its input. In simpler terms, the derivative tells us how a function's output value reacts to changes in the input value.
The process of differentiation helps us to understand the behavior of functions and is denoted by \(f'(x)\) for a given function \(f(x)\).

• Basic power rule: For any power function, \(f(x) = x^n\), the derivative is \(f'(x) = nx^{n-1}\).
• Square root function: For a function like \(\sqrt{x}\), you can apply the rule: \( \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}\).

Understanding these rules helps simplify the process of differentiation. In our example, we used basic differentiation to find the derivatives \(u'(x) = 10x\) and \(v'(x) = \frac{1}{\sqrt{x}}\), before applying the product rule. This foundational skill is necessary for analyzing and solving a wide array of mathematical problems.

Algebraic Functions

Algebraic functions are functions composed of polynomial expressions. They can include operations like addition, subtraction, multiplication, division, and taking roots. These functions often appear in basic and advanced calculus problems, making a solid understanding essential.

• Polynomial functions: These are functions of the form \(ax^n + bx^{n-1} + ... + k\). Differentiating these relies heavily on the power rule.
• Root functions: Functions such as \(x^{1/2}\) or \(\sqrt{x}\) require careful differentiation using the derivative rule for roots.

Working with algebraic functions requires combining these operations with calculus rules, like differentiation, to solve for derivatives effectively. In our example, we started with two algebraic expressions, \(5x^2 + 1\) and \(2\sqrt{x} - 1\). Understanding how to manipulate and differentiate them allowed us to successfully apply the product rule and find the overall derivative of the function. Mastering algebraic functions builds a strong foundation for tackling more complex topics in calculus.