Problem 28 - Thomas Calculus

Question

In Exercises \(19-30,\) find \(d y\) $$ y=\sec \left(x^{2}-1\right) $$

Step-by-Step Solution

Answer

\(dy = 2x \sec(x^2 - 1) \tan(x^2 - 1) dx\).

Step 1: Identify the Function

The function is given by \( y = \sec(x^2 - 1) \). Our goal is to find the differential \( dy \).

Step 2: Apply the Chain Rule

Since this is a composite function, we first find the derivative of the outer function \( \sec(u) \), and then the derivative of the inner function \( u = x^2 - 1 \).

Step 3: Differentiate the Outer Function

The derivative of \( \sec(u) \) with respect to \( u \) is \( \sec(u)\tan(u) \).

Step 4: Differentiate the Inner Function

The inner function is \( u = x^2 - 1 \). Its derivative with respect to \( x \) is \( \frac{du}{dx} = 2x \).

Step 5: Combine the Derivatives Using the Chain Rule

By the chain rule, \( \frac{dy}{dx} = \sec(u)\tan(u) \cdot \frac{du}{dx} \). Substituting, we have \( \frac{dy}{dx} = \sec(x^2 - 1)\tan(x^2 - 1) \cdot 2x \).

Step 6: Express the Differential \( dy \)

The differential is given by \( dy = \frac{dy}{dx} \cdot dx \). Substitute \( \frac{dy}{dx} \) from the previous step to get \( dy = 2x \sec(x^2 - 1)\tan(x^2 - 1)\cdot dx \).

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental tool in differential calculus, particularly useful when dealing with composite functions. It allows us to differentiate a function that is made up of two or more "linked" functions.
This technique is especially helpful because not all functions fit straightforward differentiation rules, and the chain rule bridges that gap.
Here's how it generally works:

• If you have a composite function like \( y = f(g(x)) \), the chain rule states that the derivative \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

In simpler terms, you first take the derivative of the outer function and then multiply it by the derivative of the inner function. This multiplication process gives you the derivative of the entire composite function. This rule breaks down complex problems into manageable steps.

• Think of it as peeling back layers of an onion, where each layer is a function.
• You differentiate each layer, from the outermost to the innermost.

Understanding and mastering the chain rule is vital for tackling more advanced calculus problems. It provides a strong foundation for working with any form of nested functions in mathematics.

Basics of Differentiation

Differentiation is the process by which we find the derivative of a function. Derivatives convey the rate at which a quantity changes. It's a core element of calculus with the fundamental aim of understanding how things change.
The derivative of a function at a certain point gives us the slope of the tangent line to the function at that point.
There are various rules and methods for finding derivatives:

• Power Rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
• Product Rule: For two functions \( u(x) \) and \( v(x) \), the derivative of their product is \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \).
• Quotient Rule: For a function that is the division of two functions \( \frac{u(x)}{v(x)} \), the derivative is \( \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \).

Mastering differentiation allows you to access the information about a function's behavior, such as increase or decrease, concavity, and points of inflection. It's a stepping stone to solving complex calculus problems with confidence and efficiency.

Composite Functions Explained

A composite function involves more than one function acting on an input. In mathematical terms, this means a function is nested inside another function.
To "decompose" the composite function:
Consider the expression \( y = \sec(x^2 - 1) \). Here, the function \( x^2 - 1 \) is inside the trigonometric function \( \sec(u) \).
When dealing with such functions, recognize:

• The inner function \( u(x) = x^2 -1 \).
• The outer function is \( \sec(u) \).

Breaking the problem into these components simplifies the differentiation process. First, differentiate the outer function with respect to your new variable (here \( u \)), and then the inner function with respect to the original variable (\( x \)).
A solid grasp of composite functions enables you to apply the chain rule effectively. You'll often encounter them in various math and real-world applications, making comprehension crucial for advancing in mathematics.