Problem 50 - University Calculus: Early Transcendentals

Question

Use the identity $$\csc ^{-1} u=\frac{\pi}{2}-\sec ^{-1} u$$ to derive the formula for the derivative of \(\csc ^{-1} u\) in Table 3.1 from the formula for the derivative of \(\sec ^{-1} u\)

Step-by-Step Solution

Answer

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

Step 1: Recall the given identity

The problem provides the identity \[ \csc^{-1} u = \frac{\pi}{2} - \sec^{-1} u \] This identity relates the inverse cosecant function to the inverse secant function and a constant \( \frac{\pi}{2} \).

Step 2: Differentiate both sides

We will differentiate both sides of the equation with respect to \( u \). The derivative of a constant, \( \frac{\pi}{2} \), is zero. Thus, we have:\[ \frac{d}{du} (\csc^{-1} u) = 0 - \frac{d}{du} (\sec^{-1} u) \] This simplifies to: \[ \frac{d}{du} (\csc^{-1} u) = -\frac{d}{du} (\sec^{-1} u) \]

Step 3: Recall the derivative of \(\sec^{-1} u\)

The derivative of \( \sec^{-1} u \) with respect to \( u \) is given by:\[ \frac{d}{du} (\sec^{-1} u) = \frac{1}{|u| \sqrt{u^2 - 1}} \]

Step 4: Substitute and simplify

Substitute the derivative of \( \sec^{-1} u \) into the equation from Step 2:\[ \frac{d}{du} (\csc^{-1} u) = -\frac{1}{|u| \sqrt{u^2 - 1}} \] This results in the formula for the derivative of \( \csc^{-1} u \):\[ \frac{d}{du} (\csc^{-1} u) = -\frac{1}{|u| \sqrt{u^2 - 1}} \]

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential in calculus and many areas of mathematics because they allow us to find angles when we know the value of a trigonometric function. The most common inverse trig functions include the inverse sine, cosine, tangent, cosecant, secant, and cotangent. Each function helps us solve for angles associated with various geometric and real-world problems.

When working with inverse trigonometric functions, it is crucial to remember their domains and ranges since they differ from their respective trigonometric counterparts. For example, the inverse sine function, represented as \( \sin^{-1} \) or arcsine, has a domain of \([-1, 1]\) and a range of \([-\pi/2, \pi/2]\).

Inverse trig functions can be derived from their respective trigonometric function's graphs by reflecting them over the line \( y = x \). This reflection shows us how the domain and range shift, making them navigable within calculus operations. Understanding these shifts is vital when finding derivatives, especially when simplifying expressions or integrating these functions.

Derivative of Inverse Cosecant

The derivative of the inverse cosecant function is a useful formula in calculus, especially when dealing with implicit differentiation or integrating functions involving the inverse cosecant. To derive it, we use the following identity:

• \( \csc^{-1} u = \frac{\pi}{2} - \sec^{-1} u \)
• This identity connects the inverse cosecant with the inverse secant and a constant.

Differentiating both sides with respect to \( u \) involves remembering that the derivative of a constant is 0. As a result, the differentiation process simplifies to:

\( \frac{d}{du} (\csc^{-1} u) = -\frac{d}{du} (\sec^{-1} u) \)

Knowing the derivative of the inverse secant, \( \frac{1}{|u| \sqrt{u^2 - 1}} \), allows us to substitute it directly into the equation to obtain:

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

This formula can seem daunting initially, but it's a significant tool for solving complex calculus problems, especially when inverse trigonometric functions are involved.

Derivative of Inverse Secant

The derivative of the inverse secant is pivotal in finding the derivatives of related functions, such as the inverse cosecant. The inverse secant function is represented as \( \sec^{-1} u \), and its derivative helps uncover changes in angles associated with given lengths or ratios. Here's how it sounds:

The derivative of \( \sec^{-1} u \) is determined by:

• \( \frac{d}{du} (\sec^{-1} u) = \frac{1}{|u| \sqrt{u^2 - 1}} \)
• This expression arises due to the inverse relationship with the secant's derivative and the geometric properties depicted by these functions.

To comprehend why this derivative takes this specific form, recall the Pythagorean identity \( 1 + \tan^2 \theta = \sec^2 \theta \). When differentiating an identity like \( \sec^{-1} u \), we rely on the relationships between tangent and secant functions and their inverses.

Grasping the derivative of the inverse secant equips students to handle more complicated expressions and conversions in calculus problems, enhancing problem-solving skills in various contexts.