Problem 38 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

\( \frac{dy}{ds} = \frac{s}{\sqrt{s^2 - 1}} - \frac{1}{|s|\sqrt{s^2 - 1}} \)

Step 1: Identify the terms in the function

The given function is \( y = \sqrt{s^2 - 1} - \sec^{-1}(s) \). It is composed of two terms: \( \sqrt{s^2 - 1} \) and \( -\sec^{-1}(s) \). We will find the derivative of each term separately with respect to \( s \).

Step 2: Differentiate the first term \( \sqrt{s^2 - 1} \)

The derivative of \( \sqrt{s^2 - 1} \) can be found using the chain rule. Rewrite \( \sqrt{s^2 - 1} \) as \( (s^2 - 1)^{1/2} \). The derivative \( \frac{d}{ds} (s^2 - 1)^{1/2} \) is \( \frac{1}{2}(s^2 - 1)^{-1/2} \cdot 2s = \frac{s}{\sqrt{s^2 - 1}} \).

Step 3: Differentiate the second term \( -\sec^{-1}(s) \)

The derivative of the inverse secant function is given by \( \frac{d}{ds} \sec^{-1}(s) = \frac{1}{|s|\sqrt{s^2 - 1}} \). Therefore, the derivative of \( -\sec^{-1}(s) \) is \( -\frac{1}{|s|\sqrt{s^2 - 1}} \).

Step 4: Combine the derivatives

Combine the derivatives from Steps 2 and 3. The derivative of the entire function \( y = \sqrt{s^2 - 1} - \sec^{-1}(s) \) is: \[ \frac{dy}{ds} = \frac{s}{\sqrt{s^2 - 1}} - \frac{1}{|s|\sqrt{s^2 - 1}} \].

Step 5: Simplify the expression if possible

The expression \( \frac{dy}{ds} = \frac{s}{\sqrt{s^2 - 1}} - \frac{1}{|s|\sqrt{s^2 - 1}} \) can be combined over the same denominator: \( \frac{s|s| - 1}{|s|\sqrt{s^2 - 1}} \) after factoring out the common denominator. The derivative remains as it is since this is the simplified form.

Key Concepts

Chain Rule

The chain rule is a fundamental derivative rule in calculus used for differentiating composite functions. When you have a function nested inside another, the chain rule helps calculate the derivative effectively.
For instance, in the original exercise, the term \( \sqrt{s^2 - 1} \) can be seen as a composite function where \( u = s^2 - 1 \) is inside the square root function \( u^{1/2} \). To find its derivative:

• Differentiate the outer function: If \( u = (s^2 - 1) \), then the outer function \( u^{1/2} \) has the derivative \( \frac{1}{2}u^{-1/2} \).
• Create the derivative of the inner function: \( s^2 - 1 \) differentiates to \( 2s \).
• Apply the chain rule: Multiply the derivative of the outer function by the derivative of the inner function: \( \frac{1}{2}(s^2 - 1)^{-1/2} \cdot 2s \), resulting in \( \frac{s}{\sqrt{s^2 - 1}} \).

This step-by-step process allows for accurate calculation of derivatives for complex functions.

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \sec^{-1}(s) \), offer a way to handle trigonometric values in reverse. In calculus, these functions also have specific derivatives that help analyze the rates of change associated with trigonometric influences.

• The derivative of \( \sec^{-1}(s) \) is \( \frac{1}{|s|\sqrt{s^2 - 1}} \), which includes both the absolute value \( |s| \) and a square root term \( \sqrt{s^2 - 1} \).
• When calculating the derivative of \( -\sec^{-1}(s) \), the negative sign simply factors outside the resulting expression, giving \( -\frac{1}{|s|\sqrt{s^2 - 1}} \).

Understanding these derivatives helps deal with more complex applications involving angles and their inverses, empowering you to navigate problems involving trigonometric parameters more confidently.

Calculus Problem-Solving

Solving calculus problems often involves combining various techniques and rules to find solutions to derivatives effectively. The problem-solving steps include:

• Identifying the components of the expression, as done in the first step of the solution.
• Applying derivative rules like the chain rule or inverse function derivatives precisely to each part.
• Combining the results logically, by adding or subtracting derivatives as needed.
• Simplifying the expressions, if possible, to present the cleanest form without unnecessary complexity.

For the exercise in question, the differentiation of each term was carefully conducted, and then the results were combined to provide a complete derivative of the given function. By approaching problems methodically, calculus becomes a series of manageable steps, making even intricate expressions simpler to tackle.