Problem 14 - 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 \(x .\) $$ y=\left(\frac{x}{5}+\frac{1}{5 x}\right)^{5} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \left(\frac{x}{5} + \frac{1}{5x}\right)^4 \left(1 - \frac{1}{x^2}\right) \).

Step 1: Identify Inner and Outer Functions

In the given function \( y = \left(\frac{x}{5} + \frac{1}{5x}\right)^5 \), identify \( u = g(x) = \frac{x}{5} + \frac{1}{5x} \) as the inner function. The outer function is \( f(u) = u^5 \).

Step 2: Differentiate the Inner Function

Compute \( \frac{du}{dx} \) by differentiating the function \( u = \frac{x}{5} + \frac{1}{5x} \). Using the rules of differentiation, we find: \( \frac{du}{dx} = \frac{1}{5} - \frac{1}{5x^2} \).

Step 3: Differentiate the Outer Function

Differentiate the outer function \( f(u) = u^5 \) with respect to \( u \), to find \( \frac{df}{du} = 5u^4 \).

Step 4: Apply the Chain Rule

Use the chain rule to find \( \frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \). Substitute \( \frac{du}{dx} = \frac{1}{5} - \frac{1}{5x^2} \) and \( u = \frac{x}{5} + \frac{1}{5x} \) into this expression. This gives \( \frac{dy}{dx} = 5 \left(\frac{x}{5} + \frac{1}{5x}\right)^4 \left(\frac{1}{5} - \frac{1}{5x^2}\right) \).

Step 5: Simplify the Expression

Simplify the expression for \( \frac{dy}{dx} \). This can be written as \( \frac{dy}{dx} = \left(\frac{x}{5} + \frac{1}{5x}\right)^4 \left(1 - \frac{1}{x^2}\right) \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, primarily concerned with how functions change. By finding the derivative of a function, we gain insight into its rate of change with respect to an independent variable. In the exercise provided, the function is presented in a composite form, which means we need to differentiate it carefully using rules such as the chain rule.

The goal with differentiation is to take a function and compute its derivative, represented as \( \frac{dy}{dx} \) for functions dependent on \( x \). For straightforward functions, this process involves applying basic derivative formulas. However, for composite functions, as seen in the exercise where we identify inner and outer functions, the chain rule plays a crucial role.

Inner and Outer Functions

Understanding inner and outer functions is crucial when dealing with composite functions. These are functions where one function is inside another. In our example, the function, \( y = \left(\frac{x}{5} + \frac{1}{5x}\right)^5 \), involves one function inside another, necessitating the use of the chain rule.

• The **inner function** is defined as \( u = g(x) = \frac{x}{5} + \frac{1}{5x} \).
• The **outer function**, which acts on the inner function, is \( f(u) = u^5 \).

You first differentiate the inner function to find \( \frac{du}{dx} \). Then, you differentiate the outer function with respect to \( u \) to find \( \frac{df}{du} \). Identifying these functions correctly is imperative for employing the chain rule accurately.

Simplifying Derivatives

After applying the chain rule to find the derivative, often the expression can be simplified for clarity and further analysis. Simplifying derivatives involves re-arranging and combining terms, factoring when possible, and ensuring that the derivative is presented in the cleanest form.

In the step-by-step solution:

• We conclude the process with the expression \( \frac{dy}{dx} = 5 \left(\frac{x}{5} + \frac{1}{5x}\right)^4 \left(\frac{1}{5} - \frac{1}{5x^2}\right) \). This expression is simplified further to a more elegant version: \( \left(\frac{x}{5} + \frac{1}{5x}\right)^4 \left(1 - \frac{1}{x^2}\right) \).

The simplification helps emphasize the pattern and structure within the derivative, making it easier to interpret and work with in additional applications or further computation. Simplifying is a vital step to ensure your results are not unnecessarily complex and are more user-friendly for subsequent use.