Problem 13 - Essential Calculus

Question

Find the differential \(d y\) of the given function. $$ y=\frac{1}{3} \cos \left(\frac{6 \pi x-1}{2}\right) $$

Step-by-Step Solution

Answer

The differential of the given function is \( -\pi \sin \left( \frac{6 \pi x - 1}{2} \right) \)

Step 1: Identify Function Type

First, identify that the function is a composite function, which means a function is being applied to another function. The inner function is \(\frac{6 \pi x - 1}{2}\) and the outer function is \(\frac{1}{3} \cos x\).

Step 2: Apply the Chain Rule

Next, apply the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. Differentiate the outer function with respect to its inner function to give: \( -\frac{1}{3} \sin \left( \frac{6 \pi x - 1}{2} \right) \). Then differentiate the inner function with respect to \(x\) obtaining \(3\pi\) .

Step 3: Multiply the results

Now, multiply the results from the differentiation of outer function and the inner function. This gives the differential \(d y\) of the given function: \( -3\pi \sin \left( \frac{6 \pi x - 1}{2} \right) /3 \).

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in differential calculus, especially when dealing with composite functions. When a function is applied to another function, the chain rule helps us find the derivative of the resulting composite function. It can be summarized as taking the derivative of the outer function with respect to its inner function, and then multiplying it by the derivative of the inner function with respect to the variable of interest. This technique allows us to "chain" together the derivatives of the outer and inner functions.

• Identify the outer and inner functions in the composite function.
• Differentiate the outer function with respect to the inner function.
• Differentiate the inner function with respect to the independent variable.
• Multiply these derivatives to obtain the derivative of the composite function.

This simple yet powerful rule is an excellent tool for differentiation of complex functions.

Diving Into Composite Functions

A composite function is formed when one function is placed inside another. To understand this, consider each function as a set of processes. Instead of performing one function at a time, a composite function allows for a seamless flow from one function to another. In the expression provided in the exercise, the inner function is \( \frac{6 \pi x - 1}{2} \) and the outer function is the cosine function \( \frac{1}{3} \cos(u) \). The challenge in differentiation is to handle these layered processes carefully.In practice:

• The inner function is differentiated first; this gives insight into how it changes with respect to the variable, \(x\).
• The outer function is then differentiated; however, this differentiation is with respect to its input or the inner function.

Composite functions demand a methodical approach since each layer of function builds upon the underlying one.

Differentiation Demystified

Differentiation is the process of finding the derivative of a function, which represents the rate of change of the function with respect to a variable. It is a core concept in calculus and provides vital information about the function's behavior, such as its slope at any given point. By differentiating, we can determine how fast a function is changing at a particular value of \(x\).Here’s how we approach this in any differentiation problem:

• Determine the type of function: is it simple, composite, or something else?
• Apply the appropriate differentiation rules which, in the case of composite functions, includes using the chain rule.
• Simplify the resulting expression to obtain a clean and interpretable form of the derivative.

Differentiation is not just about calculation; it involves understanding the interplay between functions and the variables they depend on. This knowledge enriches our interpretation and solution of mathematical problems.