Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The derivative is \(-\frac{2}{3}(x^3+2x)^{-5/3}(3x^2+2)\)."

Step 1: Identify the Structure of the Function

The function given is \[y = \frac{1}{(x^3 + 2x)^{2/3}}\]We recognize it as a composition of functions. It's helpful to apply the chain rule for derivatives, as there is a function inside another function. We will consider \((x^3+2x)\) as the inner function, raised to the power of \(-2/3\) since \(\frac{1}{a^n} = a^{-n}\).

Step 2: Apply the Chain Rule

We need to differentiate using the chain rule, which states that if \[y = f(g(x))\] then \[\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\]Let \[g(x) = (x^3 + 2x)\]and set \[f(u) = u^{-2/3}\] where \(u = g(x)\). Thus\[f'(u) = \frac{d}{du}u^{-2/3} = -\frac{2}{3}u^{-5/3}\]

Step 3: Compute the Derivative of the Inner Function

We find the derivative of the inner function, \[g(x) = x^3 + 2x\], by differentiating with respect to \(x\):\[g'(x) = \frac{d}{dx}(x^3+2x) = 3x^2 + 2\]

Step 4: Combine the Chain Rule Components

Substitute back to find \[\frac{dy}{dx}\]by combining the derivatives found in the previous steps:\[\frac{dy}{dx} = -\frac{2}{3} (x^3+2x)^{-5/3} \cdot (3x^2 + 2) \]Now, simplify if needed, but here it is presented in a form ready for evaluation.

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivative of a composite function. When we have a function inside another function, like nesting a simpler formula within a more complex one, the chain rule helps us differentiate it. Consider the composite function described as \( y = f(g(x)) \). The chain rule states the derivative \( \frac{dy}{dx} \) is given by the product \( f'(g(x)) \cdot g'(x) \). This means we first differentiate the outer function with respect to the inner function, and then separately differentiate the inner function with respect to \( x \), and multiply them together.

The equation used in our exercise demonstrates this brilliantly:

• First, recognize \((x^3 + 2x)\) as the inner function, \( g(x) \), raised to the power of \( -2/3 \).
• The derivative of the outer function, where \( u = g(x) \), is \( f'(u) = -\frac{2}{3}u^{-5/3} \).
• The inner function's derivative is \( g'(x) = 3x^2 + 2 \).

All these parts fit together to form the final derivative expression, showcasing the power of the chain rule in calculus.

Composite Functions

Composite functions occur when one function is applied to the result of another. This creates a new function that is essentially a combination of two functions. In symbols, if you have two functions \( f \) and \( g \), the composite function \( f(g(x)) \) means we first apply \( g \) to \( x \), and then apply \( f \) to that result.

In our exercise, the function is \( y = \frac{1}{(x^3 + 2x)^{2/3}} \). To see it as a composite function:

• The inner function \( g(x) = x^3 + 2x \) is evaluated first.
• Then, the outer function \( f(u) = u^{-2/3} \) acts on this result.

This nested behavior is central in many calculus problems, as it allows for more complex expressions to be differentiated systematically via methods like the chain rule. Recognizing composite functions is often the first step in tackling problems involving differentiation.

Derivatives of Polynomials

Differentiating polynomials is one of the most straightforward operations in calculus. Polynomials consist of terms built from variables raised to non-negative integer powers and possibly multiplied by coefficients. The basic rule for finding the derivative of a polynomial term \( ax^n \) is to multiply by the power \( n \) and reduce the power by one, resulting in \( n \cdot ax^{n-1} \).

For our function's inner part \( g(x) = x^3 + 2x \), each term is differentiated individually:

• The derivative of \( x^3 \) is \( 3x^2 \).
• The derivative of \( 2x \) is \( 2 \).

Together, these deliver \( g'(x) = 3x^2 + 2 \), which is crucial in applying the chain rule to the exercise. Understanding polynomials' derivatives serves as a foundation for dealing with more advanced calculus challenges, making it essential for students to grasp this concept fully.