Problem 229 - Calculus Volume 1

Question

Find \(\frac{d y}{d x}\) for each function. $$ y=(5-2 x)^{-2} $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = 4(5-2x)^{-3}\).

Step 1: Identify the Function Type

The given function is in the form \(y = (5-2x)^{-2}\). This is a composite function involving an outer power function and an inner linear function.

Step 2: Apply the Chain Rule

To differentiate the function \(y = (5-2x)^{-2}\), we use the chain rule. The chain rule states that if you have a function \(h(x) = f(g(x))\), then the derivative \(h'(x)\) is \(f'(g(x)) \cdot g'(x)\). Here, \(f(u) = u^{-2}\) where \(u = 5-2x\).

Step 3: Differentiate the Outer Function

The derivative of \(f(u) = u^{-2}\) with respect to \(u\) is \(f'(u) = -2u^{-3}\), by using the power rule.

Step 4: Differentiate the Inner Function

The derivative of the inner function \(g(x) = 5-2x\) is \(g'(x) = -2\). This is a simple linear function, so the derivative is the coefficient of \(x\).

Step 5: Combine Derivatives Using Chain Rule

Substituting in the derivatives into the chain rule formula, we have \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). Thus, we get:\[\frac{dy}{dx} = -2(5-2x)^{-3} \cdot (-2)\]

Step 6: Simplify the Expression

Simplify the expression by multiplying the constants:\[\frac{dy}{dx} = 4(5-2x)^{-3}\]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, representing the process of finding the derivative of a function. It essentially measures how a function changes as its input changes, i.e., the rate at which the dependent variable changes with respect to the independent variable. Differentiation allows us to determine the slope of the function at every point and can be denoted by various notations such as \(\frac{dy}{dx}\), \(y'\), or \(f'(x)\).

The basic concept of differentiation relies on applying different rules and techniques to compute derivatives effectively. Some commonly used rules include:

• Power Rule
• Product Rule
• Quotient Rule
• Chain Rule

In our example, we are interested in differentiating the function \(y = (5-2x)^{-2}\). By understanding differentiation, we can determine how quickly \(y\) changes as \(x\) changes for this specific function.

Mastering differentiation is a crucial skill in calculus because it is used extensively in physics, engineering, and economics to model real-world problems and functions.

Composite Function

A composite function is a mathematical construct where one function is composed of another function. Think of it as a chain link, where each function represents a link connected to another. If we have two functions, \(f(x)\) and \(g(x)\), the composite function is expressed as \(f(g(x))\). The value of \(f\) depends on the output of \(g\), meaning you first apply \(g\) to \(x\), and then \(f\) to the result of \(g(x)\).

In the exercise, the function \(y = (5-2x)^{-2}\) is an example of a composite function. Here,

• The inner function \(g(x) = 5-2x\)
• The outer function \(f(u) = u^{-2}\), where \(u = 5-2x\)

By recognizing the function is composite, we can effectively apply the chain rule for differentiation. Understanding how a composite function works is critical when dealing with complex functions, as it helps break down the differentiation process into manageable steps.

Power Rule

The power rule is a basic differentiation rule that makes finding derivatives straightforward when dealing with power functions. For a function of the form \(f(x) = x^n\), where \(n\) is any real number, the derivative is given by \(f'(x) = nx^{n-1}\).

In the context of our function \(y = (5-2x)^{-2}\), the power rule helps us differentiate the outer function. Here, the outer function is \(f(u) = u^{-2}\). Using the power rule, the derivative is determined as \(f'(u) = -2u^{-3}\). By reducing the exponent by one and multiplying by the original exponent, the power rule simplifies the differentiation of power functions significantly.

Applying the power rule is essential when differentiating terms raised to a power, and combining it with other differentiation rules can help tackle more complex functions efficiently.