Problem 28 - Brief Applied Calculus

Question

Find the derivative of each function. $$ f(x)=\frac{9}{2 \sqrt[3]{x^{2}}}-16 \sqrt{x^{5}}-14 $$

Step-by-Step Solution

Answer

The derivative is \(f'(x) = -3 x^{-\frac{5}{3}} - 40 x^{\frac{3}{2}}\).

Step 1: Understand the Function

The given function is \(f(x) = \frac{9}{2 \sqrt[3]{x^2}} - 16 \sqrt{x^5} - 14\). This can be rewritten using exponents as \(f(x) = \frac{9}{2} x^{-\frac{2}{3}} - 16 x^{\frac{5}{2}} - 14\). Our goal is to find the derivative \(f'(x)\).

Step 2: Use Power Rule on Each Term

For the function \(f(x) = \frac{9}{2} x^{-\frac{2}{3}} - 16 x^{\frac{5}{2}} - 14\), apply the power rule \(\frac{d}{dx}(x^n) = nx^{n-1}\) to each term. The constant \(-14\) has a derivative of \(0\).

Step 3: Derivative of First Term

Apply the power rule to the first term \(\frac{9}{2} x^{-\frac{2}{3}}\). The derivative is \(-\frac{9}{3} x^{-\frac{5}{3}} = -3 x^{-\frac{5}{3}}\).

Step 4: Derivative of Second Term

Apply the power rule to the second term \(-16 x^{\frac{5}{2}}\). The derivative is \(\frac{5}{2} \times -16 x^{\frac{3}{2}} = -40 x^{\frac{3}{2}}\).

Step 5: Combine Derivatives

Combine the results from Steps 3 and 4. Therefore, \(f'(x) = -3 x^{-\frac{5}{3}} - 40 x^{\frac{3}{2}}\).

Key Concepts

Power Rule

The Power Rule is a staple concept in calculus, integral to finding the derivatives of polynomial and non-polynomial functions. It enables us to efficiently compute the derivative of functions involving powers of a variable. This rule states that for a function of the form \( f(x) = x^n \), its derivative is given by \( f'(x) = nx^{n-1} \).

Here’s why the Power Rule is particularly useful:

• It simplifies the differentiation process of terms like \( x^2 \) or \( x^{-3} \).
• It is directly applicable to functions where the variable is raised to any real number.

In the original exercise, we used the Power Rule for each term in the rewritten function: \( \frac{9}{2} x^{-\frac{2}{3}} \) becomes \(-3 x^{-\frac{5}{3}}\) once differentiated, following this rule's straightforward application.

Derivative of Functions

Finding the derivative of a function is central in calculus and involves understanding how a function changes at any given point. The derivative gives us the rate of change or the slope of the tangent line to the curve at a point. For a function \( f(x) \), its derivative \( f'(x) \) provides insights into the increasing or decreasing nature of the graph.

When dealing with complex functions that involve multiple terms, like in the original exercise, the derivative is computed for each term separately. Common steps in this process include:

• Simplifying each function into a form that is easy to differentiate.
• Applying appropriate rules, such as the Power Rule, carefully to each term.

The derivative of simple constants, like \(-14\) in the problem, is always zero since constants do not change.

Exponentiation in Calculus

Exponentiation plays a vital role in calculus, particularly when dealing with functions expressed in terms of powers or roots. Understanding how to manipulate exponents in expressions is key to differentiating functions effectively.

Converting radical expressions to exponentials makes calculus operations easier. For instance, in the problem, \( \sqrt[3]{x^2} \) is rewritten as \( x^{-2/3} \), which aligns with applying the Power Rule. This conversion is essential because:

• It turns complex root expressions into manageable exponents.
• It helps to clearly apply differentiation rules like the Power Rule.

Being comfortable with manipulating exponents ensures that you can confidently tackle derivatives of any form, simplifying the process and reducing potential errors.