Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(d y / d x\). \(y=\left(x^{3}-2 x\right)^{1 / 3}\)

Step-by-Step Solution

Answer

\( \frac{d y}{d x} = \frac{(3x^2 - 2)}{3(x^3 - 2x)^{2/3}} \)

Step 1: Identify the Function Structure

Observe that the function is in the form of a power, specifically a cubic root. The expression inside the cubic root is \( x^3 - 2x \). Thus, we can identify that the function is of the form \( y = u^{1/3} \) where \( u = x^3 - 2x \).

Step 2: Apply the Chain Rule

To find the derivative \( \frac{d y}{d x} \), we need to apply the chain rule. The chain rule states \( \frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} \).

Step 3: Differentiate the Outer Function

The outer function is \( y = u^{1/3} \). The derivative of \( u^{1/3} \) with respect to \( u \) is \( \frac{1}{3}u^{-2/3} \). Thus, \( \frac{d y}{d u} = \frac{1}{3}u^{-2/3} \).

Step 4: Differentiate the Inner Function

The inner function is \( u = x^3 - 2x \). Differentiate \( u \) with respect to \( x \). The derivative of \( x^3 \) is \( 3x^2 \), and the derivative of \( -2x \) is \( -2 \). Therefore, \( \frac{d u}{d x} = 3x^2 - 2 \).

Step 5: Combine the Differentiated Parts

Now apply the chain rule. Substitute \( \frac{d y}{d u} = \frac{1}{3}(x^3 - 2x)^{-2/3} \) and \( \frac{d u}{d x} = 3x^2 - 2 \) into the equation: \[ \frac{d y}{d x} = \frac{1}{3}(x^3 - 2x)^{-2/3} \cdot (3x^2 - 2) \]

Step 6: Simplify the Expression

The expression for \( \frac{d y}{d x} \) becomes: \[ \frac{d y}{d x} = \frac{(3x^2 - 2)}{3(x^3 - 2x)^{2/3}} \] This is the simplified form of the derivative of the given function.

Key Concepts

Chain Rule

The chain rule is a pivotal concept in calculus, especially when dealing with composite functions. In simple terms, it allows you to differentiate a function that contains another function inside it. For example, consider a function of the form \( y = g(f(x)) \). To find \( \frac{dy}{dx} \), the chain rule states that you multiply the derivative of the outer function \( g \) with respect to \( f \), \( \frac{dg}{df} \), by the derivative of the inner function \( f \) with respect to \( x \), \( \frac{df}{dx} \). This gives us:

• \( \frac{dy}{dx} = \frac{dg}{df} \cdot \frac{df}{dx} \)

In the exercise, the function \( y = (x^3 - 2x)^{1/3} \) required the application of the chain rule. Here:

• Outer function: \( g(u) = u^{1/3} \)
• Inner function: \( u = x^3 - 2x \)

Therefore, the chain rule helps in breaking down the problem, so we can tackle each part separately and combine them to find the derivative.

Derivative

The derivative is a fundamental tool in calculus used to measure how a function changes as its input changes. In essence, the derivative provides us with the function's rate of change or its slope. For a function \( f(x) \), its derivative, denoted as \( f'(x) \) or \( \frac{df}{dx} \), tells us how \( f \) increases or decreases with a slight change in \( x \).

To compute the derivative, we use rules like the power rule, product rule, quotient rule, and the chain rule, as needed. In the exercise, after separating the inner and outer functions, we differentiated each part individually:

• For \( u = x^3 - 2x \), the derivative \( \frac{du}{dx} = 3x^2 - 2 \).
• For \( y = u^{1/3} \), the derivative \( \frac{dy}{du} = \frac{1}{3}u^{-2/3} \).

These derivatives reflect the rate of change of each respective function, guiding us to compute the overall derivative of the composite function.

Functions

Functions are the backbone of calculus, playing a central role in understanding mathematical relationships. A function, generally denoted as \( f(x) \), maps each input \( x \) to exactly one output. They come in various forms such as linear, quadratic, and exponential functions, among others. In this context, understanding how functions operate is crucial for applying differentiation.

In the exercise, the main function was expressed as \( y = (x^3 - 2x)^{1/3} \), which is a composite function formed by a cubic polynomial inside a root function. To dissect this function, we identified:

• The inner function: a cubic polynomial \( x^3 - 2x \).
• The outer function: a cubic root \( u^{1/3} \).

Decomposing this composite function was vital to using the chain rule effectively, allowing us to focus on each component separately before combining them.

Cubic Root

The cubic root is a specific type of root function, denoted as \( \sqrt[3]{x} \) or \( x^{1/3} \). It represents a value that, when multiplied by itself three times, returns the original number. The concept of roots is an inverse operation of exponents and is crucial in simplifying and transforming expressions involving powers.

In the given exercise, the outer portion of our function was the cubic root \( (x^3 - 2x)^{1/3} \). Differentiating such expressions often requires careful attention to the power rule. When applying the chain rule, the derivative of a cubic root function \( u^{1/3} \) with respect to \( u \) is calculated using the power rule:

• \( \frac{d}{du}(u^{1/3}) = \frac{1}{3}u^{-2/3} \)

Understanding how to differentiate roots effectively is key to solving a variety of calculus problems, especially those involving composite functions.