Problem 13 - Thomas Calculus

Question

In Exercises \(9-18,\) write the function in the form \(y=f(u)\) and \(u=g(x) .\) Then find \(d y / d x\) as a function of \(x .\) $$ y=\left(\frac{x^{2}}{8}+x-\frac{1}{x}\right)^{4} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 4\left( \frac{x^2}{8} + x - \frac{1}{x} \right)^3 \left( \frac{x}{4} + 1 + \frac{1}{x^2} \right) \).

Step 1: Identify Parts of the Function

First, identify the inner function \(u\) and the outer function \(f(u)\) within the given function \(y = \left( \frac{x^2}{8} + x - \frac{1}{x} \right)^4\). The inner function \(u\) is what is inside the parenthesis, while \(f(u)\) will be the expression raised to the 4th power.

Step 2: Define the Inner Function

The inner function is determined by the expression inside the parentheses: \( u = \frac{x^2}{8} + x - \frac{1}{x} \).

Step 3: Define the Outer Function

The outer function, which is \(f(u)\), takes the form of raising the inner function to the power of 4: \( y = f(u) = u^4 \).

Step 4: Differentiate the Outer Function

Apply the power rule to differentiate the outer function \( y = u^4 \) with respect to \(u\): \( \frac{dy}{du} = 4u^3 \).

Step 5: Differentiate the Inner Function

Differentiate the inner function \( u = \frac{x^2}{8} + x - \frac{1}{x} \) with respect to \(x\). This requires utilizing the rules for differentiating each term separately: \( \frac{du}{dx} = \frac{d}{dx}\left( \frac{x^2}{8} \right) + \frac{d}{dx}(x) - \frac{d}{dx}\left( \frac{1}{x} \right) \). Each term's derivative is: - \( \frac{d}{dx}\left( \frac{x^2}{8} \right) = \frac{x}{4} \), - \( \frac{d}{dx}(x) = 1 \),- \( \frac{d}{dx}\left( \frac{1}{x} \right) = -\frac{1}{x^2} \).Combine these to get: \( \frac{du}{dx} = \frac{x}{4} + 1 + \frac{1}{x^2} \).

Step 6: Apply the Chain Rule

Use the chain rule to find \( \frac{dy}{dx} \), combining the derivatives found in Steps 4 and 5: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 4u^3 \left( \frac{x}{4} + 1 + \frac{1}{x^2} \right) \).

Step 7: Substitute Back the Inner Function

Replace \(u\) with the expression \( \frac{x^2}{8} + x - \frac{1}{x} \) back into the derivative expression: \( \frac{dy}{dx} = 4\left( \frac{x^2}{8} + x - \frac{1}{x} \right)^3 \left( \frac{x}{4} + 1 + \frac{1}{x^2} \right) \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes at any point. It's like peeking under the hood of a function to see how it behaves as its input values shift slightly. When you differentiate a function, you are essentially calculating its derivative. This tells you how fast or slow it grows or shrinks as the input (usually denoted as \(x\)) changes.

Think of it like this: if you hiked up a hill, differentiation would give you the slope or steepness of the hill at every point. A steeper slope means a faster change in height as you move forward. The derivative is denoted as \( \frac{dy}{dx} \), where \(y\) represents your function. For example, if \( y = f(x) \), then the derivative \( \frac{dy}{dx} \) shows how \(y\) changes concerning \(x\).

In our exercise, differentiation is used to find how the function \( y \), which is in terms of \(x\), changes. This involves identifying and writing the parts of the function as inner and outer functions. Then, using rules such as the power rule and the chain rule to get the derivative.

Composite Function

A composite function is like a set of nested boxes or layers. It's a function that is formed by combining two functions. You can imagine it as having an inner function and then performing another function on the result of the first.

Mathematically, if you have two functions, \( f(x) \) and \( g(x) \), a composite function is indicated as \( f(g(x)) \). You start by applying \( g(x) \) and then apply \( f \) to whatever result you get from \( g(x) \).

In the provided example, the composite function is expressed through \(y = \, \left( \frac{x^2}{8} + x - \frac{1}{x} \right)^4\), where the inside, \( \frac{x^2}{8} + x - \frac{1}{x} \), acts as \(u = g(x)\), and the outer function, \(f(u)\), is \(u^4\). This nested arrangement is essential for applying the chain rule effectively during differentiation.

Power Rule

The power rule is one of the most helpful tools in calculus for finding derivatives and it's particularly simple and elegant. It says that to differentiate a function of the form \( x^n \), where \(n\) is any real number, you can bring down the exponent as a coefficient in front and decrease the exponent by one. In mathematical notation, if \( y = x^n \), then \( \frac{dy}{dx} = nx^{n-1} \).

This rule is immensely useful when dealing with polynomial terms because it simplifies the differentiation process considerably. In our exercise, the power rule is applied to the outer function \(u^4\). By bringing down the 4 as a coefficient and reducing the exponent by one, we quickly obtain \( \frac{dy}{du} = 4u^3 \).

The beauty of the power rule is in its simplicity and consistency. Whether you're dealing with a simple polynomial or part of a composite function, the power rule helps to break it down, making it manageable to find the derivative.