Problem 28 - University Calculus: Early Transcendentals

Question

Find \(d y\). $$y=\sec \left(x^{2}-1\right)$$

Step-by-Step Solution

Answer

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

Step 1: Recognize the function form

The function given is a composite function where \( y = \sec(u) \) and \( u = x^2 - 1 \). You need to use the chain rule to differentiate this function.

Step 2: Differentiate the outer function

The derivative of \( \sec(u) \) with respect to \( u \) is \( \sec(u)\tan(u) \). Therefore, the derivative of the outer function is \( \frac{d}{du}\sec(u) = \sec(u)\tan(u) \).

Step 3: Differentiate the inner function

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

Step 4: Apply the chain rule

According to the chain rule, the derivative \( \frac{dy}{dx} \) is given by \( \frac{dy}{du} \times \frac{du}{dx} \). Substitute the derivatives from Steps 2 and 3: \( \frac{dy}{dx} = \sec(u)\tan(u) \times 2x \).

Step 5: Substitute back the expression for u

Recall that \( u = x^2 - 1 \). Substitute \( u \) back into the expression to get \( \frac{dy}{dx} = \sec(x^2 - 1)\tan(x^2 - 1) \times 2x \).

Step 6: Simplify the expression

The expression for \( \frac{dy}{dx} \) simplifies to \( 2x \sec(x^2 - 1)\tan(x^2 - 1) \). Therefore, \( d y = 2x \sec(x^2 - 1)\tan(x^2 - 1) \ dx \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. It's like a mathematical assembly line where you work on each part of a function step by step. If you have a function composed of two other functions, the chain rule allows you to break down the differentiation into manageable pieces.

Here's how it works: when you have a function of the form \( y = f(g(x)) \), the chain rule states that the derivative \( \frac{dy}{dx} \) is calculated by differentiating the outer function \( f \) with respect to the inner function \( g \), then multiplying it by the derivative of the inner function \( g \) with respect to \( x \). In simpler terms:

• Differentiate the outer function, keeping the inner function intact (this gives \( \frac{dy}{du} \)).
• Multiply by the derivative of the inner function with respect to \( x \) (this gives \( \frac{du}{dx} \)).

This is essentially using the formula: \[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}. \] It's a very efficient way to manage complex differentiations.

Composite Function

A composite function is like a function inside another function. Imagine putting one function inside the other, kind of like those Russian nesting dolls. In the exercise problem, we have \( y = \sec(u) \) where \( u = x^2 - 1 \).

The composite function allows for complicated expressions to be broken down into simpler parts, making calculus tasks more manageable. You can think of it as:

• Take an inner function \( u(x) \) that modifies \( x \) first.
• Apply the outer function \( y(u) \) to the result of the inner function.

This helps in setting up for using the chain rule because you can identify the inner and outer functions easily. Recognizing a function as composite is the first step to applying the chain rule effectively.

Derivative of Trigonometric Functions

Trigonometric functions, like sine (\( \sin \)), cosine (\( \cos \)), and secant (\( \sec \)), have their own standard derivatives, which are vital when performing differentiation. In the example exercise, we needed to differentiate the secant function \( \sec(u) \).

The derivative of \( \sec(u) \) with respect to \( u \) is \( \sec(u)\tan(u) \). This is one of the standard derivatives you will often encounter. It's helpful to remember these derivatives, as they come up frequently in calculus challenges involving trigonometric functions. Some common trigonometric derivatives include:

• \( \frac{d}{dx} \sin(x) = \cos(x) \)
• \( \frac{d}{dx} \cos(x) = -\sin(x) \)
• \( \frac{d}{dx} \tan(x) = \sec^2(x) \)
• \( \frac{d}{dx} \sec(x) = \sec(x)\tan(x) \)

Being comfortable with these derivatives will simplify calculating derivatives in more intricate calculus problems.