Problem 22 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\cos ^{-1}(1 / x)$$

Step-by-Step Solution

Answer

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

Step 1: Identify the original function

The function given is \( y = \cos^{-1}(1/x) \). Our goal is to find the derivative of \( y \) with respect to \( x \).

Step 2: Understand the inverse trigonometric function derivative

We know the derivative of \( \cos^{-1}(u) \) with respect to \( u \) is \( -\frac{1}{\sqrt{1-u^2}} \). Apply this to our function with \( u = \frac{1}{x} \).

Step 3: Apply the chain rule

Use the chain rule to find \( \frac{d}{dx}[\cos^{-1}(1/x)] \). According to the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

Step 4: Differentiate the inside function

Find \( \frac{du}{dx} \) where \( u = \frac{1}{x} \). Thus, \( \frac{du}{dx} = -\frac{1}{x^2} \).

Step 5: Substitute and simplify

We have \( \frac{dy}{du} = -\frac{1}{\sqrt{1-(1/x)^2}} = -\frac{1}{\sqrt{1-1/x^2}} \). Now substitute this and \( \frac{du}{dx} \) into the chain rule: \( \frac{dy}{dx} = -\frac{1}{\sqrt{1-(1/x)^2}} \cdot (-\frac{1}{x^2}) \).

Step 6: Simplify final expression

The expression simplifies to \( \frac{1}{x^2\sqrt{1-1/x^2}} = \frac{1}{x^2\sqrt{\frac{x^2-1}{x^2}}} = \frac{x}{\sqrt{x^2-1}} \). Thus, \( \frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 1}} \).

Key Concepts

Chain rule

The chain rule is an essential method in calculus for finding the derivative of composite functions. A composite function is created when one function is inside of another, like a nesting of functions. To differentiate these, the chain rule allows us to take one step at a time, breaking the process into manageable parts. Imagine you have a function expressed as \( f(g(x)) \). The chain rule states that the derivative of this function is the outer function's derivative evaluated at the inner function times the derivative of the inner function_with respect to \( x \). This can be expressed in a formula:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

Applying the chain rule requires careful attention to detail, especially while dealing with inverse trigonometric functions like the inverse cosine in this exercise. First, we find the derivative of the outer function, then multiply by the derivative of the inner function. In other words, take it one layer at a time.Using the chain rule effectively enables the solution of complex derivative problems by simplifying them into smaller, more approachable tasks.

Inverse cosine function

The inverse cosine function, known as arccosine and denoted as \( \cos^{-1}(x) \), is the function that returns the angle whose cosine is \( x \). It is crucial in trigonometry and calculus, particularly when solving problems involving angles. The domain of the inverse cosine function is \(-1 \leq x \leq 1\), and the range is \(0 \leq y \leq \pi\) in radians. When differentiating the inverse cosine function, the result depends on the variable inside the function. The formula for the derivative of \( \cos^{-1}(u) \) with respect to \( u \) is:

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

In this exercise, the function \( y = \cos^{-1}(\frac{1}{x}) \) requires understanding that \( u \) is \( \frac{1}{x} \). Thus, when using the chain rule, you must first compute the derivative of the outer inverse cosine function before applying the derivative of \( u \) itself. This knowledge of the inverse cosine's derivative expression is pivotal for solving such exercises.

Differentiation techniques

Differentiation techniques like the chain rule, product rule, and quotient rule form the backbone of calculus, helping in finding derivatives of various types of functions. Here, we'll focus on strategies relevant to inverse trigonometric functions. These analytical techniques simplify the process of finding rates of change in complex situations.For inverse trigonometric functions like \( \cos^{-1}(x) \), it’s vital to remember specific derivative formulas and apply them accurately. These functions often require additional steps as the arguments inside may include other functions needing differentiation, making techniques such as:

• Recognizing the composition of functions.
• Understanding how to deconstruct them.
• Applying a step-by-step process for precision.

In our example with \( y = \cos^{-1}(\frac{1}{x}) \), using differentiation techniques involves first recognizing the inverse trigonometric nature of the function, then using the chain rule to account for the inner function \( \frac{1}{x} \). This requires calculating its derivative, \(-\frac{1}{x^2}\), then substituting into the chain rule formula. Such techniques make handling seemingly complex derivatives manageable and straightforward.