Problem 52 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ \tanh ^{-1}(\operatorname{coth}(x)) $$

Step-by-Step Solution

Answer

The derivative is 1.

Step 1: Recognize the Chains of Functions

The expression given is \( \tanh^{-1}(\operatorname{coth}(x)) \). Notice that this involves the inverse hyperbolic tangent function applied to the hyperbolic cotangent function. This is a composite function, allowing us to apply the chain rule. Consider \( u = \operatorname{coth}(x) \), which makes the expression \( \tanh^{-1}(u) \).

Step 2: Differentiate the Outer Function

Differentiate the outer function \( \tanh^{-1}(u) \) with respect to \( u \). Recall that the derivative is \[ \frac{d}{du} \left( \tanh^{-1}(u) \right) = \frac{1}{1-u^2}. \]

Step 3: Compute Derivative of Inner Function

Next, compute the derivative of \( u = \operatorname{coth}(x) \) with respect to \( x \). The derivative of \( \operatorname{coth}(x) \) is \[ \frac{d}{dx} \left( \operatorname{coth}(x) \right) = -\operatorname{csch}^2(x). \]

Step 4: Apply the Chain Rule

The chain rule states that the derivative of a composite function \( \tanh^{-1}(u(x)) \) is the derivative of the outer function times the derivative of the inner function. Therefore, the derivative is \[ \frac{1}{1-u^2} \times -\operatorname{csch}^2(x). \] Substituting back \( u = \operatorname{coth}(x) \), the derivative becomes \[ \frac{-\operatorname{csch}^2(x)}{1-(\operatorname{coth}(x))^2}. \]

Step 5: Simplify the Expression

To simplify, recall the identity for hyperbolic functions: \( 1 - \operatorname{coth}^2(x) = -\operatorname{csch}^2(x) \). Applying this, the expression simplifies to \[ \frac{-\operatorname{csch}^2(x)}{-\operatorname{csch}^2(x)} = 1. \] Thus, the derivative simplifies to \( 1 \).

Key Concepts

The Chain Rule in Differentiation

The derivative of a composite function may seem daunting at first, but the chain rule makes the process more manageable. When you're dealing with a function within a function, think of peeling an onion layer by layer. You'll first differentiate the outer layer and then the inner ones. Consider a composite function like \(f(g(x))\). The chain rule tells us that the derivative is the product of the derivative of \(f\) with respect to \(g\) and the derivative of \(g\) with respect to \(x\). This can be expressed as: \[( f(g(x)) )' = f'(g(x)) \cdot g'(x)\] This seamless method allows us to handle complex expressions with confidence. Always begin by identifying each function layer, and remember, practice makes perfect! The chain rule greatly simplifies differentiation challenges like the one in the original exercise. It allows us to tackle functions that might seem complex at first glance by approaching them in manageable steps.

Inverse Hyperbolic Functions Explained

Inverse hyperbolic functions may be less common but are essential elements in calculus. Like their trigonometric counterparts, these functions serve as inverses to hyperbolic functions. For instance, \(\tanh^{-1}(x)\) is the inverse of the hyperbolic tangent function, and they are often used in hyperbolic identities involving exponential functions. Key characteristics of inverse hyperbolic functions include: - They are useful in solving equations involving hyperbolic functions. - They extend the relationship between arcs and sectors in hyperbolic space. - They follow distinct derivative rules, such as: \[\frac{d}{dx}(\tanh^{-1}(x)) = \frac{1}{1-x^2}\] In the original problem, \(\tanh^{-1}\) serves as the outer function in our composite, showcasing how inverses can transform complex expressions into solvable forms. Understanding these functions allows us to expand our calculus skills beyond the basics, offering insights into higher-level mathematics and real-world applications.

Differentiating Composite Functions

Differentiation of composite functions involves a structured approach to manage layered expressions. The original exercise showcases a composite of two hyperbolic functions, requiring us to efficiently use the chain rule. First, identify the composition within the functions. In the problem, \(\tanh^{-1}(\operatorname{coth}(x))\) comprises an outer function \(\tanh^{-1}\) and an inner function \(\operatorname{coth}(x)\). By writing \(u = \operatorname{coth}(x)\), the composite can be rewritten for clarity. Next, differentiate individually: - The derivative of the outer function, \(\tanh^{-1}(u)\), was found to be \(\frac{1}{1-u^2}\). - The derivative of the inner function, \(\operatorname{coth}(x)\), is \(-\operatorname{csch}^2(x)\). Finally, apply the chain rule:\[\frac{d}{dx}(\tanh^{-1}(\operatorname{coth}(x))) = \frac{1}{1-(\operatorname{coth}(x))^2} \cdot (-\operatorname{csch}^2(x))\] Upon simplifying, these steps reveal how each layer complements the other, turning a complex problem into an accessible solution. Mastery of composite function differentiation unravels the deeper understanding of calculus essentials, paving the way for tackling diverse mathematical challenges.