Problem 10 - Brief Applied Calculus

Question

Find the derivative of each function by using the Product Rule. Simplify your answers. $$ f(x)=6 \sqrt[3]{x}(2 x+1) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 16x^{1/3} + 2x^{-2/3} \).

Step 1: Identify Parts of the Product

The function given is a product of two functions: \( u(x) = 6\sqrt[3]{x} \) and \( v(x) = (2x + 1) \). We will apply the Product Rule, which states: \( (uv)' = u'v + uv' \).

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

To find \( u'(x) \), differentiate \( 6\sqrt[3]{x} \). We can rewrite \( \sqrt[3]{x} \) as \( x^{1/3} \), so \( u(x) = 6x^{1/3} \). The derivative is \( u'(x) = 6 \cdot \frac{1}{3}x^{-2/3} = 2x^{-2/3} \).

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

The function \( v(x) = 2x + 1 \) is linear, so its derivative is simple. Applying basic differentiation rules, we get \( v'(x) = 2 \).

Step 4: Apply the Product Rule

Using the product rule \( (uv)' = u'v + uv' \), substitute the parts: \( u'(x) = 2x^{-2/3} \), \( u(x) = 6x^{1/3} \), \( v(x) = (2x + 1) \), \( v'(x) = 2 \). Substituting these, the derivative is:\[ f'(x) = (2x^{-2/3})(2x + 1) + (6x^{1/3})(2) \].

Step 5: Simplify the Expression

First simplify each term separately:- \( 2x^{-2/3}(2x + 1) = 4x^{1/3} + 2x^{-2/3} \) (Distribute \( 2x^{-2/3} \) to each term in \( (2x + 1) \)).- \( 6x^{1/3} \cdot 2 = 12x^{1/3} \).Combine the terms: \( f'(x) = 4x^{1/3} + 2x^{-2/3} + 12x^{1/3} \). Simplify the like terms, \( 4x^{1/3} + 12x^{1/3} = 16x^{1/3} \).Final simplified derivative: \[ f'(x) = 16x^{1/3} + 2x^{-2/3} \].

Key Concepts

Derivative

The concept of a derivative is central to calculus. It represents the rate at which a function is changing at any given point. To differentiate a function means to find its derivative. Understanding derivatives is crucial because they allow us to analyze how functions behave.

For example, consider the function in this exercise, which is a product of two other functions. To find its derivative, we first identify each component of the product. In our function, we have two parts, namely, \( u(x) = 6\sqrt[3]{x} \) and \( v(x) = (2x + 1) \). Differentiation involves identifying these parts and using rules to find their individual derivatives.

Simplification

Simplification is about making expressions easier to understand or work with. After deriving a function, it's essential to simplify it for clarity and ease of use. Simplification can involve combining like terms or reducing complex expressions to simpler forms.

In the solution provided, once we find the derivative using the Product Rule, we end up with terms like \( 4x^{1/3} \), \( 2x^{-2/3} \), and \( 12x^{1/3} \). By combining the like terms \( 4x^{1/3} \) and \( 12x^{1/3} \), we simplify the expression to a neater form \( 16x^{1/3} + 2x^{-2/3} \). This makes the result more manageable and understandable.

Simplifying derivatives is not just about aesthetics; it also helps in further applications and analysis of the function.

Differentiation Rules

Differentiation rules guide us on how to find the derivative of various types of functions. These rules include the power rule, product rule, quotient rule, and chain rule, among others. In this exercise, we focus on the Product Rule.

The Product Rule is applied when you have a function that is the product of two other functions. The rule states: \((uv)' = u'v + uv'\). It provides a systematic way to differentiate products by considering the derivative of each component and how they interact.

To apply the Product Rule effectively, begin by differentiating each component separately. In this case, we differentiate \( u(x) = 6x^{1/3} \) to get \( u'(x) = 2x^{-2/3} \) and \( v(x) = 2x + 1 \) to get \( v'(x) = 2 \).

Then, by plugging these into the Product Rule formula, we handle the more complex task of differentiating the entire function. Mastering differentiation rules enables us to tackle a broad spectrum of problems involving changing rates and trends.