Question

For the following exercises, find the derivatives tor the functions.\(\ln \left(\tanh ^{-1}(x)\right)\)

Short answer

The derivative is \( \frac{1}{\tanh^{-1}(x)(1-x^2)} \).

Step-by-step solution

Differentiate the Outer Function

The function we need to differentiate is \( f(x) = \ln \left( \tanh^{-1}(x) \right) \). Start by differentiating the outer function, \( \ln(u) \), where \( u = \tanh^{-1}(x) \). The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \).

Differentiate the Inner Function

Now, differentiate the inner function \( u = \tanh^{-1}(x) \) with respect to \( x \). The derivative of \( \tanh^{-1}(x) \) is \( \frac{1}{1-x^2} \).

Apply the Chain Rule

Using the chain rule, combine the derivatives from Step 1 and Step 2. Multiply the derivative of the outer function by the derivative of the inner function: \( \frac{1}{\tanh^{-1}(x)} \times \frac{1}{1-x^2} \).

Simplify the Expression

The expression \( \frac{1}{\tanh^{-1}(x) (1-x^2)} \) is simplified as the derivative of \( \ln(\tanh^{-1}(x)) \).

Key concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus, particularly when dealing with composite functions. It allows us to differentiate functions that are nested within each other by taking the derivative of each component separately. Imagine you have a function that looks like "an onion", where layers are stacked within each other.

In our exercise, we are given a function of the form \( f(x) = \ln(\tanh^{-1}(x)) \). Here, \( \ln(u) \) acts as our outer function and \( \tanh^{-1}(x) \) is the inner function, \( u \). To apply the chain rule, you would first differentiate the outer function, treating the inner function as if it were a simple variable. Then, differentiate the inner function itself with respect to \( x \).

Here's a step-by-step breakdown of the process:

• Differentiate the outer function: The derivative of \( \ln(u) \) is \( \frac{1}{u} \), where \( u = \tanh^{-1}(x) \).
• Differentiate the inner function: The derivative of \( \tanh^{-1}(x) \) is \( \frac{1}{1-x^2} \).
• Multiply the results: Combine both derivatives by multiplying them: \( \frac{1}{\tanh^{-1}(x)} \times \frac{1}{1-x^2} \).

This systematic approach helps in finding the overall derivative while maintaining the structure of the function.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are analogous to inverse trigonometric functions though they deal with hyperbolic functions like \( \tanh, \sinh, \cosh \), etc.

In the exercise \( \tanh^{-1}(x) \) is the inverse hyperbolic tangent, which reverses the operations of the \( \tanh(x) \) function. These functions are used when dealing with problems involving hyperbolic behavior, such as physical models or certain calculus problems.

For our specific function, understanding the inverse hyperbolic tangent is crucial:

• The inverse hyperbolic tangent \( \tanh^{-1}(x) \) is defined for \(-1 < x < 1\). This domain is important because it ensures a real output.
• The derivative of \( \tanh^{-1}(x) \) is \( \frac{1}{1-x^2} \). Recognizing this derivative helps in simplifying the process when applying the chain rule. It explains how \( \tanh^{-1}(x) \) changes as \( x \) changes.

Mastering inverse hyperbolic functions can lead to an easier application of many calculus operations, especially when linked with differentiation rules.

Logarithmic Differentiation

Logarithmic differentiation is a technique used when dealing with complex functions, particularly those involving products, quotients, or powers of functions. Although the exercise doesn’t explicitly require this method, understanding it enriches one’s toolbox in calculus.

This technique simplifies differentiation by taking the natural logarithm (\( \ln \)) of both sides of an equation, turning products into sums and quotients into differences due to logarithmic properties. Logarithmic differentiation is especially useful:

• When your function includes complicated products of functions: By using logarithms, multiplication becomes addition, which is easier to differentiate.
• When having exponentials: It simplifies differentiation where the exponent is a variable.

In our specific problem, taking the logarithm of an inverse hyperbolic function already involves the natural logarithm function, which we differentiated using the chain rule. However, understanding this broader technique offers powerful strategy insights for tackling sophisticated calculus problems.