Problem 50 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ \arcsin (\tanh (x)) $$

Step-by-Step Solution

Answer

The derivative of \( \arcsin(\tanh(x)) \) is \( \text{sech}(x) \).

Step 1: Recognize the Composition of Functions

The given expression \( \arcsin(\tanh(x)) \) is a composite function. It can be denoted as \( f(g(x)) \) where \( f(u) = \arcsin(u) \) and \( g(x) = \tanh(x) \). To differentiate it, we will use the chain rule.

Step 2: Apply the Chain Rule

The chain rule states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \). Therefore, we first need to find \( f'(u) \) and \( g'(x) \).

Step 3: Differentiate the Outer Function

The outer function is \( f(u) = \arcsin(u) \). The derivative of \( \arcsin(u) \) with respect to \( u \) is \( f'(u) = \frac{1}{\sqrt{1 - u^2}} \).

Step 4: Differentiate the Inner Function

The inner function is \( g(x) = \tanh(x) \). The derivative of \( \tanh(x) \) with respect to \( x \) is \( g'(x) = \text{sech}^2(x) \). Recall that \( \text{sech}(x) = \frac{1}{\cosh(x)} \).

Step 5: Substitute and Simplify

Substitute \( g'(x) \) and \( f'(g(x)) \) into the chain rule formula. We have:\[\frac{d}{dx} \arcsin(\tanh(x)) = \frac{1}{\sqrt{1 - (\tanh(x))^2}} \cdot \text{sech}^2(x)\]Simplify using the identity \( 1 - \tanh^2(x) = \text{sech}^2(x) \), so:\[\frac{d}{dx} \arcsin(\tanh(x)) = \frac{\text{sech}^2(x)}{\sqrt{\text{sech}^2(x)}} = \text{sech}(x)\]Thus, the derivative simplifies to \( \text{sech}(x) \).

Key Concepts

Chain Rule

The chain rule is an essential concept in calculus, widely used to find the derivative of composite functions. A composite function is one where a function is applied to the result of another function, expressed as \( f(g(x)) \). To differentiate this, the chain rule tells us:

• Find the derivative of the outer function \( f \), using the inner function \( g(x) \) as its variable.
• Multiply this derivative by the derivative of the inner function \( g(x) \).

This rule allows the handling of complex expressions in a systematic and step-by-step manner. For instance, in the given problem with \( \arcsin(\tanh(x)) \), we treated \( \arcsin(u) \) as the outer function and \( \tanh(x) \) as the inner function. Applying the chain rule helped in finding the derivative by taking the derivative of \( \arcsin \) with respect to \( u \), and \( \tanh \) with respect to \( x \). This process simplifies our work considerably when dealing with layered functions.

Composite Function

A composite function is a function within a function, represented as \( f(g(x)) \). It comprises two or more functions where the output of one function becomes the input of another. For example, in \( \arcsin(\tanh(x)) \), \( \tanh(x) \) acts as the input to the \( \arcsin \) function.

• Inner Function: \( g(x) = \tanh(x) \)
• Outer Function: \( f(u) = \arcsin(u) \)

Working with composite functions can be intuitive once you recognize which parts of an expression are 'nested.' When you're tackling a differentiation problem involving these types of functions, identifying which part is inside another helps in applying the chain rule effectively. Recognizing composite functions is crucial in calculus as they frequently appear in both simple and complex mathematical contexts.

Hyperbolic Functions

Hyperbolic functions, such as \( \sinh \), \( \cosh \), and \( \tanh \), are analogs of trigonometric functions but relate to hyperbolas rather than circles. They have similar properties and derivative rules that resemble their trigonometric counterparts.

• The hyperbolic tangent function, \( \tanh(x) \), is defined as \( \frac{\sinh(x)}{\cosh(x)} \).
• Its derivative is \( \text{sech}^2(x) \), where \( \text{sech}(x) \) is the hyperbolic secant, expressed as \( \frac{1}{\cosh(x)} \).

Understanding hyperbolic functions and their derivatives is essential when dealing with certain calculus problems. They form the core of many industries, including engineering and physics, allowing more accurate models and computations within hyperbolic geometry.

Inverse Trigonometric Functions

Inverse trigonometric functions reverse the roles of the input and output of regular trigonometric functions. They find an angle when given a ratio. In our exercise, \( \arcsin \) is used, which finds the angle whose sine is a given number. In calculus, these functions have specific derivative formulas:

• For \( \arcsin(u) \), the derivative is \( \frac{1}{\sqrt{1-u^2}} \) with respect to \( u \).
• This derivative expression helps in handling functions involving arcsine by simplifying how the change in the angle is influenced by other variables.

These inverse functions often appear when solving calculus problems, as they provide a method to transition between the angle measurement and trigonometric ratios, offering solutions in analytical math and geometry.