Problem 36 - Thomas Calculus

Question

In Exercises \(25-36,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=\cosh ^{-1}(\sec x), \quad 0 < x < \pi / 2$$

Step-by-Step Solution

Answer

The derivative is \( \sec x \).

Step 1: Understand the Inverse Hyperbolic Function

The function given is \( y = \cosh^{-1}(\sec x) \). The inverse hyperbolic function \( \cosh^{-1}(x) \) is defined for \( x \geq 1 \), which aligns with \( \sec x = 1/\cos x \) being greater than or equal to 1 for \( 0 < x < \pi/2 \). This means the problem is defined appropriately in this interval.

Step 2: Differentiate \( y \) with Respect to \( \sec x \)

Recall that the derivative of \( y = \cosh^{-1}(u) \) with respect to \( u \) is \( \frac{1}{\sqrt{u^2 - 1}} \). Applying this, the derivative of \( y = \cosh^{-1}(\sec x) \) with respect to \( \sec x \) is \( \frac{1}{\sqrt{\sec^2 x - 1}} \).

Step 3: Simplify the Expression

We know that \( \sec^2 x - 1 = \tan^2 x \) based on trigonometric identities. Therefore, \( \sqrt{\sec^2 x - 1} = \sqrt{\tan^2 x} = |\tan x| \). Since \( \tan x > 0 \) in \( 0 < x < \pi/2 \), we have \( \sqrt{\sec^2 x - 1} = \tan x \).

Step 4: Use the Chain Rule

To find \( \frac{dy}{dx} \), use the chain rule: \( \frac{dy}{dx} = \frac{dy}{d(\sec x)} \cdot \frac{d(\sec x)}{dx} \). We already found \( \frac{dy}{d(\sec x)} = \frac{1}{\tan x} \). For \( \frac{d(\sec x)}{dx} = \sec x \tan x \).

Step 5: Combine the Derivatives

Multiply \( \frac{1}{\tan x} \) by \( \sec x \tan x \), which gives \( \frac{dy}{dx} = \sec x \). Thus, the derivative of \( y \) with respect to \( x \) is \( \sec x \).

Key Concepts

Inverse Hyperbolic Functions

Inverse hyperbolic functions are analogs of the trigonometric functions but for hyperbolas, rather than circles. In this exercise, we deal with the inverse hyperbolic cosine, written as \( \cosh^{-1}(x) \). This function is defined for \( x \geq 1 \) since the range of \( \cosh(x) \) is \([1, \infty)\). Understanding its domain is crucial when differentiating, as you need to ensure that \( x \) fits within this range.
In the original exercise, the function \( y = \cosh^{-1}(\sec x) \) makes use of \( \sec x \), which is always \( \geq 1 \) in the interval \( 0 < x < \pi/2 \). Thus, it aligns with the domain of \( \cosh^{-1}(x) \).
Remember, the derivative of \( \cosh^{-1}(u) \) regarding \( u \) itself involves an important expression: \( \frac{1}{\sqrt{u^2 - 1}} \). This formula arises due to its relation to the hyperbolic identity \( \cosh^2(x) - \sinh^2(x) = 1 \). It is this identity that helps us simplify and solve differentiation problems involving hyperbolic functions.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. This rule allows you to take the derivative of a function composed of other functions. In essence, if you have a function \( y = f(g(x)) \), the chain rule states that the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
In our specific example, the function \( y = \cosh^{-1}(\sec x) \) requires the use of the chain rule. Here, \( g(x) = \sec x \) and \( f(u) = \cosh^{-1}(u) \). Hence, we first differentiate \( y \) with respect to \( \sec x \), and then multiply it by the derivative of \( \sec x \) with respect to \( x \).

• The derivative of \( y = \cosh^{-1}(u) \) with respect to \( u \) is \( \frac{1}{\sqrt{u^2 - 1}} \).
• The derivative of \( \sec x \) with respect to \( x \) is \( \sec x \tan x \).

By applying the chain rule, we effectively combine these derivatives to solve the differentiation problem.

Trigonometric Identities

Trigonometric identities are essential tools for simplifying expressions and solving equations in trigonometry and calculus. They relate the various trigonometric functions to one another. A crucial identity used in this problem is \( \sec^2 x - 1 = \tan^2 x \), which stems from the identity \( \sec^2 x = 1 + \tan^2 x \).
Knowing this identity can help simplify expressions significantly. For example, in our step-by-step solution, when we have \( \sqrt{\sec^2 x - 1} \), we recognize it as \( \sqrt{\tan^2 x} \), simplifying to \( \tan x \) because \( \tan x > 0 \) within the interval \( 0 < x < \pi/2 \).
These trigonometric identities simplify the differentiation processes by reducing complex expressions to simpler ones, allowing us to find derivatives more efficiently. Mastering these identities can be incredibly beneficial not only in calculus but in various applications of mathematics.