Problem 12 - Calculus - AP Edition

Question

$$\text { Find } f^{\prime}(x)$$ $$f(x)=7 x^{-6}-5 \sqrt{x}$$

Step-by-Step Solution

Answer

\( f'(x) = -42x^{-7} - \frac{5}{2}x^{-1/2} \)

Step 1: Rewrite the Function

First, rewrite the function in a form that makes it easier to differentiate. The given function is \( f(x) = 7x^{-6} - 5\sqrt{x} \). Recall that \( \sqrt{x} = x^{1/2} \), so the function can be rewritten as \( f(x) = 7x^{-6} - 5x^{1/2} \).

Step 2: Differentiate Using Power Rule

Apply the power rule for differentiation, which states that if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \). Start by differentiating each term of \( f(x) = 7x^{-6} - 5x^{1/2} \).

Step 2.1: Differentiate the First Term

Using the power rule on the first term, \( 7x^{-6} \), gives: \( f'(x) = 7(-6)x^{-6-1} = -42x^{-7} \).

Step 2.2: Differentiate the Second Term

Using the power rule on the second term, \( -5x^{1/2} \), gives: \( f'(x) = -5(1/2)x^{(1/2)-1} = -\frac{5}{2}x^{-1/2} \).

Step 3: Combine the Derivative Terms

Combine the derivatives of the terms to find the derivative of the entire function. Thus, \( f'(x) = -42x^{-7} - \frac{5}{2}x^{-1/2} \).

Key Concepts

Power Rule

The Power Rule is a fundamental principle in differentiation that simplifies finding derivatives. This rule states: if you have a function of the form \( f(x) = x^n \), the derivative \( f'(x) \) is calculated as \( nx^{n-1} \). This transformation makes handling powers of \( x \) straightforward.Let's take an example based on the given problem:

• The function \( f(x) = 7x^{-6} - 5x^{1/2} \) involves terms with varying powers of \( x \).
• For the power \(-6\): Apply the Power Rule to \( 7x^{-6} \), giving the derivative \( -42x^{-7} \).
• For the power \( 1/2 \): Apply the Power Rule again to \( -5x^{1/2} \), giving the derivative \( -\frac{5}{2}x^{-1/2} \).

With the Power Rule, each term is differentiated individually, using the power and multiplying it by the existing coefficient of the term.

Derivative

A derivative represents the rate of change of a function with respect to one of its variables. In this context, a derivative shows how a function changes at any given point.Why Derivatives are Important:

• They enable us to understand the behavior and slope of functions.
• They help in finding maxima and minima, critical points in graph plotting.
• Derivatives are essential in fields like physics for finding velocity and acceleration.

The derivative \( f'(x) = -42x^{-7} - \frac{5}{2}x^{-1/2} \) describes how the original function \( f(x) = 7x^{-6} - 5x^{1/2} \) changes at each point \( x \). It gives precise insight into the function's slope at the desired points.

Calculus

Calculus is a branch of mathematics that explores change and motion handling through derivatives and integrals. It provides the tools to model and understand the dynamics of systems in real and theoretical environments.Why Calculus is Fundamental:

• Calculus underpins much of advanced math and science, explaining rates and trends.
• It helps solve complex equations that involve changes, like those found in physics and engineering.
• It lays the foundation for problem-solving in real-life situations, involving variable rates and flows.

In our case, the use of differentiation to find \( f'(x) = -42x^{-7} - \frac{5}{2}x^{-1/2} \) is a direct application of calculus principles. It showcases the power of calculus in simplifying and understanding complex functions through derivative computation.