Problem 281 - Calculus Volume 1

Question

For the following exercises, find \(\frac{d y}{d x}\) for the given function. $$y=\sec ^{-1}\left(\frac{1}{x}\right)$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = -\frac{x}{\sqrt{1-x^2}} \).

Step 1: Understand the Function

The given function is an inverse trigonometric function: \[ y = \sec^{-1}\left(\frac{1}{x}\right) \]The goal is to find \( \frac{dy}{dx} \), the derivative of \( y \) with respect to \( x \).

Step 2: Use Inverse Function Derivative Formula

The derivative of \( \sec^{-1}(u) \) with respect to \( u \) is given by:\[ \frac{d}{du}[\sec^{-1}(u)] = \frac{1}{|u|\sqrt{u^2-1}} \] Now, apply this formula where \( u = \frac{1}{x} \).

Step 3: Find \( \frac{du}{dx} \)

Since \( u = \frac{1}{x} \), differentiate \( u \) with respect to \( x \): \[ \frac{du}{dx} = \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \]

Step 4: Apply the Chain Rule

By the chain rule, \( \frac{dy}{dx} = \frac{d}{du}[\sec^{-1}(u)] \cdot \frac{du}{dx} \). Substitute the expressions found:\[ \frac{dy}{dx} = \frac{1}{|u|\sqrt{u^2-1}} \cdot \frac{du}{dx} \]Substitute back \( u = \frac{1}{x} \) and \( \frac{du}{dx} = -\frac{1}{x^2} \).

Step 5: Substitute and Simplify

First substitute \( u = \frac{1}{x} \): \[ \frac{dy}{dx} = \frac{1}{\left|\frac{1}{x}\right|\sqrt{\left(\frac{1}{x}\right)^2-1}} \times \left(-\frac{1}{x^2}\right) \]Calculate the absolute value and simplify:\[ \frac{dy}{dx} = -\frac{x}{\sqrt{1-x^2}} \]

Key Concepts

Derivative of Inverse Functions

To find the derivative of inverse trigonometric functions, it's important to recognize their unique nature. These functions, like \(\sec^{-1}(x)\,\), represent angles whose trigonometric ratio equals the given value. When finding derivatives, a special set of formulas is used, different from the usual trigonometric derivatives. For the secant inverse function, the derivative with respect to its argument \(u\) is:

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

Here,

• \(u\) represents a substitution inside our desired function.

So, if \(y = \sec^{-1}\left(\frac{1}{x}\right)\), we substitute \(u = \frac{1}{x}\) in the formula. Ultimately, understanding that each inverse trigonometric function has a specific derivative formula is key. Always start by identifying which inverse function you're dealing with and use the appropriate formula.
Recognize that the domain and range restrictions are essential—this affects the value of \(u\) and helps ensure you're working within the correct boundaries.

Chain Rule in Calculus

The chain rule is a fundamental concept in calculus used when differentiating compositions of functions. For a function that is composed of another function, such as \(y = \sec^{-1}\left(\frac{1}{x}\right),\) the chain rule allows us to effectively find the derivative.To apply the chain rule:

• First, differentiate the outer function while leaving the inner function untouched.
• Then, multiply by the derivative of the inner function.

In this problem, \(u = \frac{1}{x}\) is the inner function. Thus,

• the derivative of the outer function (\(\sec^{-1}(u)\)) is \(\frac{1}{|u|\sqrt{u^2-1}}\).
• The derivative of the inner function (\(\frac{1}{x}\)) is \(-\frac{1}{x^2}\).

The chain rule combines these results, so that:

• \(\frac{dy}{dx} = \frac{1}{|u|\sqrt{u^2-1}} \cdot \frac{-1}{x^2}\)

This process requires careful substitution and simplification to reach the final derivative.
Mastering the chain rule can significantly simplify complex differentiation tasks and is essential for handling composite functions smoothly.

Secant Function Derivative

Understanding the derivative of the secant function and its inverse is crucial in solving problems involving these functions. For the regular secant function (\(\sec(x)\)), its derivative is:

• \(\frac{d}{dx}[\sec(x)] = \sec(x)\tan(x)\)

This differs significantly from the derivative of the inverse secant function,\( \sec^{-1}(x)\,\). The process involves using the inverse function derivative formula, as described.The derivatives of the secant inverse require understanding of both the positive and negative values of \(x\), represented by the absolute value.

• In this exercise, \(\sec^{-1}\left(\frac{1}{x}\right)\) flips the regular process, telling us to consider \(\frac{dy}{dx} = -\frac{x}{\sqrt{1-x^2}}\).

This highlights how inverse trigonometric functions behave differently due to their geometric origin. The inverse secant derivative encapsulates the slope of the angle with respect to its input and requires careful handling, particularly around values that make the original trigonometric function undefined.