Problem 15 - Essential Calculus Early Transcendentals

Question

Prove that \(\frac{d}{d x}\left(\csc ^{-1} x\right)=-\frac{1}{x \sqrt{x^{2}-1}}\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We are asked to find the derivative of the inverse cosecant function, \( \csc^{-1}(x) \), and show that it equals \( -\frac{1}{x\sqrt{x^2 - 1}} \).

Step 2: Use Inverse Function Derivative Formula

Recall that if \( y = \csc^{-1}(x) \), then \( x = \csc(y) \). The derivative of an inverse function follows the rule \( \frac{d}{dx}\left( g^{-1}(x) \right) = \frac{1}{g'(y)} \). We first need to find \( \frac{dy}{dx} \).

Step 3: Find Derivative of Original Function

Since \( x = \csc(y) = \frac{1}{\sin(y)} \), we will differentiate \( x \) with respect to \( y \). Using chain rule and differentiation, we find:\[ \frac{dx}{dy} = -\csc(y)\cot(y) = -x\cot(y) \]

Step 4: Express \( \cot(y) \) in Terms of \( x \)

From the identity \( \cot^2(y) = \csc^2(y) - 1 \), substitute \( \csc(y) = x \):\[ \cot^2(y) = x^2 - 1 \] \[ \cot(y) = \sqrt{x^2 - 1} \]

Step 5: Substitute and Simplify

Now substitute \( \cot(y) = \sqrt{x^2 - 1} \) in \( -x\cot(y) \), so:\[ \frac{dx}{dy} = -x\sqrt{x^2 - 1} \] Then, \( \frac{dy}{dx} = \frac{-1}{x\sqrt{x^2 - 1}} \)

Step 6: Conclude with the Final Result

Thus, the derivative of \( \csc^{-1}(x) \) is\[ \frac{d}{dx}(\csc^{-1}(x)) = -\frac{1}{x\sqrt{x^2 - 1}} \] which matches the expression we needed to prove.

Key Concepts

Cosecant Function

The cosecant function, abbreviated as "csc," is one of the six main trigonometric functions. The cosecant of an angle in a right triangle is the length of the hypotenuse divided by the length of the opposite side. In mathematical terms, for an angle \( \theta \), the cosecant is defined as: \( \csc(\theta) = \frac{1}{\sin(\theta)} \). This means that cosecant is the reciprocal of the sine function.

For values of \( \theta \) that result in sine being zero, the cosecant function is undefined, making it important to consider the domain of \( \csc(\theta) \). Typically, it is defined for all real numbers \( \theta \), except where \( \sin(\theta) = 0 \).

This concept of the cosecant function extends to inverse trigonometric functions, leading us to functions like \( \csc^{-1}(x) \), which help in finding angles given values of cosecant."

Derivative of Inverse Functions

Inverse functions reverse the operation of a function. When dealing with inverse trigonometric functions like \( \csc^{-1}(x) \), we often need to find their derivatives. The derivative of an inverse function \( g^{-1}(x) \) can be found using the formula: \( \frac{d}{dx}[g^{-1}(x)] = \frac{1}{g'(y)} \).

Here, \( g(y) \) is the original function, and \( y \) is such that \( g^{-1}(x) = y \). This implies \( x = g(y) \), and gives us an essential link to find the derivative of the inverse function.

In context, finding the derivative of \( \csc^{-1}(x) \) involves considering the equation \( x = \csc(y) \). By differentiating \( x \) with respect to \( y \) and using the relationship provided by the inverse function derivative formula, we are able to deduce the correct derivative form.

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and are true for any angle. These identities are fundamental in simplifying trigonometric expressions. One commonly used identity helps relate secant and tangent: \( \cot^2(y) = \csc^2(y) - 1 \).

This identity becomes crucial when expressing one trigonometric function in terms of another. In our exercise to find \( \frac{d}{dx}(\csc^{-1}(x)) \), it enables us to express the cotangent function in terms of cosecant, leading to \( \cot(y) = \sqrt{x^2 - 1} \). This substitution simplifies the differentiation process and aligns with the established identities.

Utilizing such identities is key in overcoming algebraic complexities in differentiation tasks.

Differentiation Techniques

Differentiation is the mathematical process of finding a function's derivative, which measures how a function changes as its input changes. Different techniques are applied depending on the function being differentiated. For inverse trigonometric functions, it is essential to know and apply the correct formulas and identities.

In this exercise, we used the chain rule, which is a technique for differentiating compositions of functions. For example, when differentiating \( x = \csc(y) \), recognizing the inter-relation between \( x \) and \( y \) via \( \csc(y) \) helps us apply the chain rule effectively.

After finding \( \frac{dx}{dy} \), the reciprocal \( \frac{dy}{dx} \) provides us with the final differential form for \( \csc^{-1}(x) \). It's about transforming the problem into understandable parts and using dependable rules to solve it efficiently.