Problem 25 - Thomas Calculus

Question

In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\sec ^{-1}(2 s+1) $$

Step-by-Step Solution

Answer

The derivative is \( \frac{2}{|2s + 1|\sqrt{4s^2 + 4s}} \).

Step 1: Recall the Definition of the Arcsecant Derivative

The derivative of the inverse secant function, \( \sec^{-1}(u) \), with respect to \( u \) is \( \frac{1}{|u|\sqrt{u^2 - 1}} \). We will use this to find the derivative of \( y = \sec^{-1}(2s + 1) \) with respect to \( s \).

Step 2: Identify the Inside Function

In the expression \( y = \sec^{-1}(2s + 1) \), the inside function is \( u = 2s + 1 \). We'll need the derivative of this inside function as part of our process.

Step 3: Differentiate the Inside Function

Differentiate \( u = 2s + 1 \) with respect to \( s \). Since \( u = 2s + 1 \), its derivative \( \frac{du}{ds} \) is simply \( 2 \).

Step 4: Apply the Chain Rule

Use the chain rule: \( \frac{dy}{ds} = \frac{dy}{du} \cdot \frac{du}{ds} \). From Step 1, \( \frac{dy}{du} = \frac{1}{|u|\sqrt{u^2 - 1}} \). Combine this with \( \frac{du}{ds} = 2 \) from Step 3.

Step 5: Substitute Back the Inside Function

Substitute \( u = 2s + 1 \) into the derivative formula: \( \frac{dy}{du} = \frac{1}{|2s + 1|\sqrt{(2s + 1)^2 - 1}} \).

Step 6: Simplify the Expression

Complete the differentiation by computing the full expression: \( \frac{dy}{ds} = \frac{2}{|2s + 1|\sqrt{(2s + 1)^2 - 1}} \). Simplify \((2s + 1)^2 - 1 \) to get \( 4s^2 + 4s + 1 - 1 = 4s^2 + 4s \).

Step 7: Final Simplification

The final derivative \( \frac{dy}{ds} = \frac{2}{|2s + 1|\sqrt{4s^2 + 4s}} \) can be simplified if needed, but this is the complete derivative.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse operations of the standard trigonometric functions. They allow us to find an angle given a trigonometric ratio. Common inverse trigonometric functions include: arc sine, arc cosine, arc tangent, and arc secant, among others. The notation often used is a superscript "+1", such as in \( \sec^{-1}(x) \), which denotes the arcsecant function.
For calculus and differentiation, inverse trigonometric functions typically have rational formulas. Understanding these derivatives is crucial when dealing with problems involving compositions of functions with inverse trigonometric components.

Chain Rule

The chain rule is a fundamental technique in calculus for differentiating composite functions. When a function is composed of two other functions, the derivative is found by multiplying the derivative of the external function by the derivative of the internal function.
In symbolic terms, if a function \(y\) is composed of two functions, such that \(y=f(g(x))\), then the derivative is given by the formula:

• \( \frac{dy}{dx} = \frac{dy}{dg} \cdot \frac{dg}{dx} \)

This rule is especially vital when dealing with inverse trigonometric functions, as often they are nested within other algebraic expressions.

Arcsecant Function

The arcsecant function, denoted as \( \sec^{-1}(x) \), is the inverse of the secant function. It maps a value back to an angle whose secant is the given value. The derivative of the arcsecant function is particularly important in calculus.
The standard derivative formula for the arcsecant function \( \sec^{-1}(u) \) with respect to \(u\) is:

• \( \frac{d}{du} \sec^{-1}(u) = \frac{1}{|u|\sqrt{u^2 - 1}} \)

This formula is used frequently in problems involving differentiation of inverse trigonometric functions.

Differentiation

Differentiation is the process of computing the derivative of a function, which tells us how the function values change with respect to changes in the input. It is one of the two main operations in calculus, the other being integration.
The derivative of a function gives the slope of the tangent to the graph of the function at any point. In this particular exercise, differentiating \(y = \sec^{-1}(2s + 1)\) involves recognizing the internal function \(u = 2s + 1\) and applying the chain rule to find the derivative with respect to \(s\).
This involves calculating:

• The derivative of the internal function \( \frac{du}{ds} \)
• The derivative of the inverse function \( \frac{dy}{du} \)
• Combining these results to find \( \frac{dy}{ds} \)

Through understanding these steps, mastery of the derivative computation using rules of differentiation becomes attainable.