Question

Find \(D_{x} y\). $$ y=\operatorname{coth}^{-1}(\tanh x) $$

Short answer

\( D_{x} y = \operatorname{sech}(x) \).

Step-by-step solution

Understand the hyperbolic functions involved

The hyperbolic cotangent inverse function, \( \operatorname{coth}^{-1}(x) \), and the hyperbolic tangent function, \( \tanh(x) \), are the main functions involved here. We need to understand their basic properties for differentiation.

Identify the composition of functions

We have the function \( y = \operatorname{coth}^{-1} (u) \) where \( u = \tanh(x) \). Therefore, this is a composition of functions and will require the chain rule to differentiate.

Differentiate using the Chain Rule

To find \( \frac{dy}{dx} \), apply the chain rule. First, differentiate \( y \) with respect to \( u \): \( \frac{dy}{du} = -\frac{1}{u^2 - 1} \) since \( \operatorname{coth}^{-1}(u) = \frac{1}{2} \ln\left( \frac{u+1}{u-1} \right) \).

Differentiate \( u = \tanh(x) \)

Differentiate \( u = \tanh(x) \) with respect to \( x \) to get \( \frac{du}{dx} = \operatorname{sech}^2(x) \).

Apply the Chain Rule

Apply the chain rule \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = -\frac{1}{u^2 - 1} \times \operatorname{sech}^2(x) \). Substitute \( u = \tanh(x) \) into the equation.

Simplify

Since \( u = \tanh(x) \), we know \( u^2 = \tanh^2(x) \), and \( u^2 - 1 = \tanh^2(x) - 1 = -\operatorname{sech}^2(x) \). Now substitute everything back into \( \frac{dy}{dx} \). This gives \( \frac{dy}{dx} = \operatorname{sech}(x) \).

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but are based on hyperbolas rather than circles. The hyperbolic tangent function, \( \tanh(x) \), and the hyperbolic cotangent inverse function, \( \operatorname{coth}^{-1}(x) \), are the two key players here. These functions help model phenomena in many fields, like engineering and physics.

Just like their trigonometric counterparts, hyperbolic functions fulfill certain identities. For instance:

• The identity \( \tanh^2(x) + \operatorname{sech}^2(x) = 1 \).
• The derivative of \( \tanh(x) \) is \( \operatorname{sech}^2(x) \).

These properties are used to find derivatives involving hyperbolic functions.

The function \( y = \operatorname{coth}^{-1}(\tanh(x)) \) involves both the \( \tanh \) function and its inverse hyperbolic function, \( \operatorname{coth}^{-1} \), providing a slightly more intricate case for differentiation.

Chain Rule

The chain rule is an essential tool in calculus. It helps us differentiate composite functions—functions within functions.

Think of the chain rule as peeling an onion. You deal with each layer or function separately. In the problem above, you first encounter \( \operatorname{coth}^{-1}(u) \) where \( u = \tanh(x) \). This setup fits the precise scenario ideal for applying the rule.

When using the chain rule, follow these steps:

• Differentiate the outer function \( y = \operatorname{coth}^{-1}(u) \) with respect to \( u \) to get \( \frac{dy}{du} = -\frac{1}{u^2-1} \).
• Then, differentiate the inner function \( u = \tanh(x) \) with respect to \( x \), resulting in \( \frac{du}{dx} = \operatorname{sech}^2(x) \).

Finally, combine these derivatives using the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \].
This method efficiently unravels the composite function for differentiation.

Differentiation Techniques

Differentiation is about finding how quickly a function changes. It’s crucial in understanding dynamic systems. Various techniques exist, and selecting the right one can simplify complex problems.

For the current problem, you leverage the chain rule, a central differentiation technique, to unravel the nested functions. However, understanding the specific derivatives of hyperbolic functions is equally important. Remember that:

• The derivative of \( \tanh(x) \) results in \( \operatorname{sech}^2(x) \).
• The expression \( \operatorname{coth}^{-1}(u) \) differentiates into a rational function \( \frac{-1}{u^2-1} \).

Emphasizing these derivatives, the chain rule simplifies \( \frac{dy}{dx} \) from an expression that looks complex to one that is reduced and simplified: \[ \frac{dy}{dx} = \operatorname{sech}(x) \].
Hence, using the appropriate techniques can make differentiation not only feasible but straightforward.