Problem 26 - Thomas Calculus

Question

In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\sec ^{-1} 5 s $$

Step-by-Step Solution

Answer

The derivative is \(\frac{1}{|s|\sqrt{25s^2-1}}\).

Step 1: Recall the Derivative Formula for Inverse Secant

The derivative of the inverse secant function is given by:\[\frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{|x|\sqrt{x^2-1}}\]

Step 2: Apply the Chain Rule

Since the given function is \(y = \sec^{-1}(5s)\), use the chain rule to find the derivative. Let \(u = 5s\), then \(y = \sec^{-1}(u)\). Find \(\frac{dy}{du}\) and \(\frac{du}{ds}\).

Step 3: Differentiate the Outer Function

Differentiate the function \(y = \sec^{-1}(u)\) with respect to \(u\):\[\frac{dy}{du} = \frac{1}{|u|\sqrt{u^2-1}}\]

Step 4: Differentiate the Inner Function

Differentiate \(u = 5s\) with respect to \(s\):\[\frac{du}{ds} = 5\]

Step 5: Combine Using the Chain Rule

Apply the chain rule to find \(\frac{dy}{ds}\):\[\frac{dy}{ds} = \frac{dy}{du} \times \frac{du}{ds} = \frac{1}{|5s|\sqrt{(5s)^2-1}} \times 5\]Simplify to obtain:\[\frac{dy}{ds} = \frac{5}{|5s|\sqrt{25s^2-1}}\]

Step 6: Simplify the Expression

The result from the previous step is already simplified for the derivative of \(y\) with respect to \(s\). The final answer is:\[\frac{dy}{ds} = \frac{1}{|s|\sqrt{25s^2-1}}\]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. When you have a function nested within another function, the chain rule allows you to find the derivative of the entire expression efficiently.
The general formula for the chain rule is:

• If you have a composite function \( y = f(g(x)) \), then the derivative is given by \( \frac{dy}{dx} = f'(g(x)) \, g'(x) \)

In simpler terms, you first differentiate the outer function while keeping the inner function intact, then multiply by the derivative of the inner function.
For example, in the exercise where \( y = \sec^{-1}(5s) \), the outer function \( y = \sec^{-1}(u) \) and the inner function \( u = 5s \). By applying the chain rule, you found the derivatives of each and combined them to obtain the final derivative \( \frac{dy}{ds} \). This method is extremely useful across many areas of calculus where functions are layered together.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as the inverse secant \( \sec^{-1}(x) \), are the inverses of the standard trigonometric functions. They allow us to work backwards from the ratio of sides of a right triangle to the angle itself.
These functions are useful in situations where we need to determine an angle given a particular trigonometric ratio.
Each inverse function has a specific derivative formula:

• For \( \sec^{-1}(x) \), the derivative is \( \frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{|x|\sqrt{x^2-1}} \)

This derivative formula reflects the rate of change of the angle with respect to changes in the secant ratio. In the provided exercise, this formula was used when finding \( \frac{dy}{du} \), after setting \( u = 5s \).
Inverse trigonometric derivatives are common in calculus problems, especially when dealing with angles and their relationships in calculus and geometry applications.

Differentiation

Differentiation is the process of finding the derivative of a function, a crucial concept in calculus. The derivative provides the rate at which a function is changing at any given point and is fundamental in understanding and analyzing the behavior of functions.
Differentiation can be performed on various types of functions, including polynomial, exponential, logarithmic, and trigonometric functions. For each type of function, specific rules and formulas are applied to find its derivative efficiently.

• The power rule, product rule, quotient rule, and chain rule are some of the basic tools used in differentiation.

In the exercise, differentiation was central to finding \( \frac{dy}{ds} \), where you computed the derivative of the composite function \( \sec^{-1}(5s) \) with respect to \( s \).
This involved differentiating both the inner and outer functions separately and then combining them using the chain rule, demonstrating the interconnectedness of these differentiation tools.