Question

Prove the formula for the derivative of \(y=\cosh ^{-1}(x)\) by differentiating \(x=\cosh (y)\). (Hint: Use hyperbolic trigonometric identities.)

Short answer

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

Step-by-step solution

Express Inverse Function Relationship

The hyperbolic cosine inverse function is defined by the relationship \( x = \cosh(y) \). So we have \( y = \cosh^{-1}(x) \). We will first express \( x \) in terms of \( y \).

Differentiate Both Sides

Differentiate \( x = \cosh(y) \) with respect to \( y \) to get \( \frac{dx}{dy} = \sinh(y) \). Here, \( \sinh(y) \) is the derivative of \( \cosh(y) \).

Solve for \( \frac{d\sinh(y)}{dy} \)

To obtain \( \frac{dy}{dx} \), we need to find \( \frac{dy}{dx} = \frac{1}{\sinh(y)} \). Since \( \sinh(y) = \sqrt{\cosh^2(y) - 1} \), and \( x = \cosh(y) \), we conclude \( \sinh(y) = \sqrt{x^2 - 1} \).

Substitute Expression for \( \sinh(y) \)

Now substitute this expression into the derivative relation: \( \frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 1}} \).

Verify Result with Hyperbolic Identity

Use the identity \( \cosh^2(y) - \sinh^2(y) = 1 \) to ensure that the relation \( \sinh(y) = \sqrt{x^2 - 1} \) holds by substitution back, confirming the manipulation steps are valid.

Key concepts

Inverse Functions and Their Importance

Inverse functions allow us to reverse the effects of an original function. Essentially, they "undo" what the original function does. For example, if the function is multiplication by 2, the inverse would be division by 2. This concept helps us solve equations and understand relationships between variables more deeply.

In mathematics, inverse functions are often signified by a superscript -1. Just like in our problem, the inverse of the hyperbolic function \( \cosh(y) \) is \( \cosh^{-1}(x) \). In the context of hyperbolic functions, the inverse function allows us to express \( y \) in terms of \( x \), or vice versa. This connection is essential for derivative calculation, because knowing how to flip the function relationship lets us tackle derivative problems from the opposite angle.

• Inverse functions are used to solve for variables by reversing operations.
• They're essential for derivative calculations as they allow changing the variable of differentiation.
• Notation such as \( f^{-1}(x) \) is used to represent inverse functions.

Derivative Calculation: Breaking Down the Process

When you hear the term 'derivative', think of it as a way to find the rate at which one quantity changes with respect to another. It’s like asking, "How fast is something happening?"

For our hyperbolic function \( x = \cosh(y) \), we differentiate with respect to \( y \) to find how \( x \) changes for small changes in \( y \). The derivative of \( \cosh(y) \) is \( \sinh(y) \), giving us the key expression \( \frac{dx}{dy} = \sinh(y) \).

To find \( \frac{dy}{dx} \), we need the inverse operation: instead of checking how \( x \) changes as \( y \) varies, we look for the rate of change of \( y \) as \( x \) changes. Using the relation \( \sinh(y) = \sqrt{x^2 - 1} \), we switch our perspective thanks to functions being invertible, leading us to \( \frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 1}} \).

• Derivatives express the concept of change or rate.
• Essential for optimizing, minima, and maxima problems.
• Use differentiation rules to handle complex functions, like hyperbolics.

Exploring Hyperbolic Trigonometric Identities

Hyperbolic trigonometric identities are to hyperbolic functions what ordinary trigonometric identities are to regular trigonometric functions. They provide necessary shortcuts and essential tools for simplifying complex mathematical expressions.

When dealing with hyperbolic functions like \( \cosh(y) \), recognizing identities helps us rewrite expressions more conveniently. One key identity used here is \( \cosh^2(y) - \sinh^2(y) = 1 \). These identities deep-link different hyperbolic functions, simplifying calculations and derivations.

In our derivation of the inverse hyperbolic function \( y = \cosh^{-1}(x) \), using the identity \( \sinh(y) = \sqrt{\cosh^2(y) - 1} \) was crucial. It allowed us to express \( \sinh(y) \) in terms of \( x \), facilitating the conversion from \( \frac{dx}{dy} \) to \( \frac{dy}{dx} \).

• Hyperbolic trigonometric identities simplify complex expressions.
• They are crucial for derivations involving hyperbolic functions.
• They show relationships between \( \cosh, \sinh \), and others that aid in calculations.