Question

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

Short answer

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

Step-by-step solution

Understand the Problem

We need to find the derivative of the inverse hyperbolic secant function, which is written as \( y = \operatorname{sech}^{-1}(x) \). We start by considering its inverse function relation: \( x = \operatorname{sech}(y) \).

Express \( \operatorname{sech}(y) \) in Terms of Exponentials

Recall that the hyperbolic secant is defined as \( \operatorname{sech}(y) = \frac{2}{e^y + e^{-y}} \). From the given \( x = \operatorname{sech}(y) \), we have \( x = \frac{2}{e^y + e^{-y}} \).

Implicitly Differentiate Both Sides

Differentiate both sides with respect to \( x \), using implicit differentiation on the right side. Since \( x = \operatorname{sech}(y) \), we have \( 1 = \frac{d}{dx}\left(\frac{2}{e^y + e^{-y}}\right) \).

Apply Chain Rule

Apply the chain rule to differentiate the right side: \[ \frac{d}{dx} \left(x\right) = \frac{d}{dy} \left(\frac{2}{e^y + e^{-y}}\right) \cdot \frac{dy}{dx} \]The derivative of \( \operatorname{sech}(y) \) with respect to \( y \) is \(-\operatorname{sech}(y)\operatorname{tanh}(y) \). Thus: \[ 1 = -\operatorname{sech}(y)\operatorname{tanh}(y) \cdot \frac{dy}{dx} \]

Solve for \( \frac{dy}{dx} \)

Now solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{1}{-\operatorname{sech}(y)\operatorname{tanh}(y)} \]

Express \( \operatorname{sech}(y) \) and \( \operatorname{tanh}(y) \) in terms of \( x \)

Since \( x = \operatorname{sech}(y) \), we have:\[ \operatorname{tanh}(y) = \sqrt{1 - \operatorname{sech}^2(y)} = \sqrt{1 - x^2} \]Thus, \[ \frac{dy}{dx} = \frac{1}{-x \sqrt{1 - x^2}} \]

Finalize the Derivative Expression

Therefore, the derivative of \( y = \operatorname{sech}^{-1}(x) \) is:\[ \frac{dy}{dx} = -\frac{1}{x \sqrt{1 - x^2}} \]

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the trigonometric functions but are based on hyperbolas rather than circles. The hyperbolic secant, denoted as \( \operatorname{sech}(y) \), is defined by the formula \( \operatorname{sech}(y) = \frac{2}{e^y + e^{-y}} \). This means it relates exponential functions in a way similar to how trigonometric functions relate angles to unit circles.

There are a few key hyperbolic functions to be familiar with:

• Hyperbolic Sine (\(\sinh(x)\)): Defined as \(\sinh(x) = \frac{e^x - e^{-x}}{2}\). It's similar to the normal sine function but stretches across the hyperbola.
• Hyperbolic Cosine (\(\cosh(x)\)): Defined as \(\cosh(x) = \frac{e^x + e^{-x}}{2}\). It models the shape of a hanging cable or chain, known as a catenary.
• Hyperbolic Tangent (\(\tanh(x)\)): It’s the ratio \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\).

These functions have unique properties and are used in various fields such as engineering and physics to model scenarios that go beyond simple angles in a plane.

Understanding these functions and their properties is crucial for dealing with problems that involve inverse hyperbolic functions. They provide the backbone for differentiating and integrating inverse hyperbolic functions as seen in problems like proving derivatives.

Implicit Differentiation

Implicit differentiation is a technique used when you have an equation \(y\) defined implicitly in terms of \(x\). Unlike explicit differentiation where \(y\) is explicitly solved in terms of \(x\), implicit differentiation handles situations where separating them isn't feasible. It allows for finding the derivative with respect to \(x\) for equations that may not be solved explicitly for one variable.

Here's how it works:

• Start with an equation involving both \(x\) and \(y\).
• Differentiating each term of the equation on both sides with respect to \(x\), applying relevant derivative rules.
• For terms involving \(y\), apply the chain rule: differentiate \(y\) as if \(y\) were \(x\), then multiply by \(\frac{dy}{dx}\).
• Solve the resulting equation to find \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\).

In the provided exercise, implicit differentiation was used to find the derivative of \(\operatorname{sech}(y)\) with respect to \(x\), ultimately aiming to derive \(\frac{dy}{dx}\). This step is essential when you deal with inverse functions or any implicitly defined relationships.

Derivative Formulas

Derivative formulas are essential tools in calculus, allowing us to find the rate of change of functions. When dealing with inverse hyperbolic functions, understanding the derivative rules is vital. This involves the application of standard derivative formulas and the chain rule.

For instance, using the derivative of the hyperbolic secant, \( \operatorname{sech}(y) = \frac{2}{e^y + e^{-y}} \), we apply implicit differentiation:

• First, express \(\operatorname{sech}(y) \) using its exponential form.
• Differentiate with respect to \(y\), getting \(-\operatorname{sech}(y)\operatorname{tanh}(y)\).
• Apply chain rule to establish the relationship between \(\frac{dy}{dx}\) and the differential \(-\operatorname{sech}(y)\operatorname{tanh}(y)\).

When proving the derivative of an inverse function, like \( y = \operatorname{sech}^{-1}(x) \), derivative formulas are applied to transform the implicit differentiation result into an explicit \(\frac{dy}{dx}\) expression.

Being adept with these derivative techniques helps solve complex calculus problems efficiently. They provide the framework necessary for exploring more nuanced relationships between functions, such as the one found in hyperbolic and inverse trigonometric functions.