Problem 38 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\sqrt{s^{2}-1}-\sec ^{-1} s$$

Step-by-Step Solution

Answer

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

Step 1: Differentiate the First Term

The first term is \(\sqrt{s^2-1}\). We can rewrite it as \((s^2 - 1)^{1/2}\). Use the chain rule for differentiation: let \(u = s^2 - 1\), then \(\sqrt{u} = u^{1/2}\). The derivative of \(u^{1/2}\) with respect to \(u\) is \(\frac{1}{2}u^{-1/2}\), and the derivative of \(u\) with respect to \(s\) is \(2s\). Therefore, the derivative of \(\sqrt{s^2-1}\) is \(\frac{1}{2}(s^2 - 1)^{-1/2} \cdot 2s = \frac{s}{\sqrt{s^2-1}}\).

Step 2: Differentiate the Second Term

The second term is \(\sec^{-1}(s)\). The derivative of \(\sec^{-1}(s)\) with respect to \(s\) is \(\frac{1}{|s|\sqrt{s^2-1}}\). Keep in mind the absolute value in the formula to handle both positive and negative values of \(s\).

Step 3: Combine the Derivatives

Now, combine the results from Steps 1 and 2. Since the function \(y\) is given by a subtraction of two terms, \(\frac{dy}{ds} = \frac{s}{\sqrt{s^2-1}} - \frac{1}{|s|\sqrt{s^2-1}}\).

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus that is used to find the derivative of a function that is composed of other functions. If you have a function of the form \(f(g(x))\), where \(f\) and \(g\) are both differentiable, the chain rule states that the derivative of this function is \(f'(g(x)) \, g'(x)\). This allows us to tackle more complex problems by breaking them into manageable parts.

• Identify the inner and outer functions. The inner function is the function composed within another, and the outer function is applied to the results of the inner function.
• Differentiate the outer function with respect to the inner function first, and then multiply by the derivative of the inner function.

In the original exercise, we used the chain rule to differentiate \((s^2 - 1)^{1/2}\). Here, \(u = s^2 - 1\) is the inner function and \(u^{1/2}= \sqrt{u}\) is the outer. So, the derivative becomes \(\frac{1}{2}(s^2 - 1)^{-1/2} \times 2s\). This simplification is a classic application of the chain rule.

Trigonometric Functions

Trigonometric functions derive from the relationships between the angles and lengths of right triangles. They are periodic and have derivatives that are also trigonometric functions. Common examples include sine, cosine, and tangent, which are fundamental in mathematics and applied sciences.

• These functions have specific properties and are often used in waves and oscillations.
• They have well-known derivatives, for instance, the derivative of \(\sin(x)\) is \(\cos(x)\), and the derivative of \(\cos(x)\) is \(-\sin(x)\).

In terms of derivatives, trigonometric functions are particularly important because they require specific rules when differentiating more complex forms, such as the composite functions highlighted in the chain rule section.

Inverse Trigonometric Derivatives

Inverse trigonometric functions, such as \(\sec^{-1}(x)\), help to find an angle given a trigonometric ratio. Their derivatives are essential for calculus due to their unique properties.

• These derivatives often involve expressions with an absolute value or root, as inverse trig functions can account for multiple quadrant angles.
• For example, the derivative of \(\sec^{-1}(s)\) is \(\frac{1}{|s|\sqrt{s^2-1}}\), which shows its complexity.

Understanding these derivatives allows for advanced problem-solving capabilities in calculus. They truly exemplify how you can remove the trigonometric component from complex problems, leaving algebraic manipulation for solutions. In the original problem, the derivative of \(\sec^{-1}(s)\) showcases the added layer of thought required when dealing with absolute values and roots in differential equations.