91Ó°ÊÓ

In calculus, implicit differentiation is a technique used to find the derivative of a variable with respect to another variable when they are linked by an implicit function. An implicit function is not expressed explicitly in terms of one variable, such as \( y = f(x) \), but rather in a mixed form like \( F(x, y) = 0 \).

Here, we wish to determine the derivatives of the inverse hyperbolic functions \( ext{sinh}^{-1}(x), ext{cosh}^{-1}(x), ext{and tanh}^{-1}(x) \). These derivatives are found by working through a process involving implicit differentiation of the hyperbolic functions.

• For \( ext{sinh}^{-1}(x) \), start with \( x = ext{sinh}(y) \). Differentiating implicitly with respect to \( x \) involves treating \( y \) as a function of \( x \) and applying the chain rule.
• Repeating this logic, apply implicit differentiation to \( x = ext{cosh}(y) \) and \( x = ext{tanh}(y) \) to find the derivatives for \( ext{cosh}^{-1}(x) \), and \( ext{tanh}^{-1}(x) \).

Implicit differentiation involves several layers of understanding as it integrates multiple mathematical concepts. Each step requires exact calculations, often using the chain rule, to isolate the derivative of \( y \) with respect to \( x \).

The chain rule is a fundamental tool in calculus to determine the derivative of composite functions. This rule states that if a variable \( z \) depends on a variable \( y \), which in turn depends on another variable \( x \), then \( z \) also depends on \( x \). The relationship can be expressed as

• \( \frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx} \)

In our exercise involving inverse hyperbolic functions, the chain rule is essential when differentiating functions that inherently possess a nested structure.

For instance, when starting with the relation \( x = ext{sinh}(y) \), implicitly differentiate to get:

• \( \frac{d}{dx}x = \frac{d}{dx}( ext{sinh}(y)) \)
• Applying the chain rule yields \( 1 = ext{cosh}(y) \cdot \frac{dy}{dx} \)

The chain rule allows us to account for how y changes as x changes when y itself is a function involving hyperbolic functions. This is similarly applied to the other exercises with \( x = ext{cosh}(y) \) and \( x = ext{tanh}(y) \), leading to their respective derivatives.

Hyperbolic functions, like \( ext{sinh}, ext{cosh}, ext{and tanh} \), are analogues of the trigonometric functions but for the hyperbola. They play a key role in many branch of calculus and are defined as follows:

• \( ext{sinh}(x) = \frac{e^x - e^{-x}}{2} \)
• \( ext{cosh}(x) = \frac{e^x + e^{-x}}{2} \)
• \( ext{tanh}(x) = \frac{ ext{sinh}(x)}{ ext{cosh}(x)} \)

Their derivatives are equally important and are used in both implicit differentiation and the chain rule.

• The derivative of \( ext{sinh}(x) \) is \( ext{cosh}(x) \).
• The derivative of \( ext{cosh}(x) \) is \( ext{sinh}(x) \).
• The derivative of \( ext{tanh}(x) \) is \( ext{sech}^2(x) \), where \( ext{sech}(x) = \frac{1}{ ext{cosh}(x)} \).

These derivatives are pivotal when finding how \( y \) changes in relation to \( x \) in each of the original exercises. Hypothetical calculus problems involving hyperbolic functions often require a solid grasp of their properties and derivatives to accurately solve them.