Problem 8 - Thomas Calculus

Question

Find \(d y / d x\) in Exercises \(1-10\) $$ y=(1-6 x)^{2 / 3} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = -4 (1 - 6x)^{-1/3} \).

Step 1: Identify the function and its form

The given function is \( y = (1 - 6x)^{2/3} \). This is in the form of a composition of functions, where the outer function is a power function and the inner function is linear. This indicates that we need to use the chain rule for differentiation.

Step 2: Apply the Chain Rule

The chain rule states that if you have a composite function \( y = f(g(x)) \), its derivative is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). Here, \( f(u) = u^{2/3} \) and \( g(x) = 1 - 6x \).

Step 3: Differentiate the Outer Function

First, differentiate the outer function with respect to \( u \). For \( f(u) = u^{2/3} \), the derivative is \( f'(u) = \frac{2}{3}u^{-1/3} \).

Step 4: Differentiate the Inner Function

Now differentiate the inner function \( g(x) = 1 - 6x \) with respect to \( x \). The derivative is \( g'(x) = -6 \).

Step 5: Combine the Derivatives

Apply the chain rule: \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). Replace \( u \) with \( g(x) \) to have: \( \frac{dy}{dx} = \frac{2}{3}(1 - 6x)^{-1/3} \cdot (-6) \).

Step 6: Simplify the Expression

Simplify the derivative calculation: \( \frac{dy}{dx} = \frac{2}{3} \cdot (-6) \cdot (1 - 6x)^{-1/3} = -4 (1 - 6x)^{-1/3} \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus used to find the rate at which a quantity changes. Essentially, it helps us understand how a function behaves when its input changes slightly.

In the exercise given, our task is to determine the derivative of the function. The derivative tells us the rate of change of the dependent variable with respect to the independent variable. In simpler terms, it describes how fast or slow the function's output value changes as the input value changes.

• The derivative of a function, often denoted as \( \frac{dy}{dx} \), provides an equation that allows us to compute the slope of the tangent line at any point on the curve of the function.
• This slope represents the steepness or inclination of the curve at a particular point and tells us whether the function is increasing or decreasing at that point.
• Understanding differentiation is key to solving real-world problems involving rates of change, such as speed, growth, decay, and optimization issues.

With practice, applying differentiation becomes a powerful tool for analyzing and interpreting a wide range of mathematical models.

Composite Functions

Composite functions are created when one function is applied to the result of another function. In the context of the chain rule, composite functions consist of an "outer" function and an "inner" function.

For our function \( y = (1 - 6x)^{2/3} \), we see that it's composed of two parts:

• The **inner function** is \( g(x) = 1 - 6x \).
• The **outer function** is \( f(u) = u^{2/3} \), where \( u = g(x) \).

Composite functions require careful handling during differentiation because the change in the whole function depends on both inner and outer parts. Using the chain rule is very effective here.

The chain rule allows us to differentiate composite functions efficiently by breaking them down into their components and differentiating each in turn before assembling the result. Understanding how to recognize and differentiate composite functions is crucial for solving more complex calculus problems efficiently.

Power Function Differentiation

Power functions are functions of the form \( y = x^n \), where "n" is any real number. Differentiation of power functions is straightforward, thanks to the power rule.

The power rule states that if \( y = x^n \), then the derivative \( \frac{dy}{dx} = nx^{n-1} \). This rule makes it easy to find the rate of change of power functions.

In our exercise's composite function \( y = (1 - 6x)^{2/3} \), we differentiate the outer part, \( u^{2/3} \), using the power rule:

• We consider \( u = (1 - 6x) \) as a single entity.
• Differentiating \( u^{2/3} \) gives us \( \frac{2}{3} u^{-1/3} \).

This process exemplifies how the power rule is applied in more complex situations when combined with the chain rule.

Mastering power function differentiation is fundamental for tackling various calculus tasks, from simple polynomial functions to advanced calculus involving more layers of complexity.