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Magazine Chapter 6: Problem 24
Find \(d y / d x\). $$y=\left(\operatorname{coth}^{-1} x\right)^{2}$$
Short Answer
Expert verified
\( \frac{dy}{dx} = -\frac{2 \operatorname{coth}^{-1}(x)}{1-x^2} \)
Step by step solution
01
Apply the Chain Rule
We have a composition of functions, where \( y = u^2 \) and \( u = \operatorname{coth}^{-1}(x) \). First, apply the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \] We will calculate \( \frac{dy}{du} \) and \( \frac{du}{dx} \) separately in the next steps.
02
Differentiate the Outer Function
The outer function is \( u^2 \) with respect to \( u \). Differentiate it:\[ \frac{dy}{du} = 2u \]Since \( u = \operatorname{coth}^{-1}(x) \), \( \frac{dy}{du} = 2 \operatorname{coth}^{-1}(x) \).
03
Differentiate the Inner Function
The inner function is \( u = \operatorname{coth}^{-1}(x) \). The derivative of \( \operatorname{coth}^{-1}(x) \) with respect to \( x \) is:\[ \frac{du}{dx} = -\frac{1}{1-x^2} \] This formula assumes that \( |x| > 1 \).
04
Combine Results Using the Chain Rule
Substitute the results from Step 2 and Step 3 into the chain rule from Step 1:\[ \frac{dy}{dx} = 2 \operatorname{coth}^{-1}(x) \cdot \left(-\frac{1}{1-x^2}\right) \]Simplify the expression:\[ \frac{dy}{dx} = -\frac{2 \operatorname{coth}^{-1}(x)}{1-x^2} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate compositions of functions. When you have a function nested within another function, like in the expression \( y=(\operatorname{coth}^{-1} x)^{2} \), the chain rule assists you in finding the derivative by breaking down the differentiation process into manageable parts.
- Identify the Outer and Inner Functions: In our problem, the outer function is \( u^2 \), where \( u = \operatorname{coth}^{-1} x \), and the inner function is the inverse hyperbolic function \( \operatorname{coth}^{-1} x \).
- Differentiate Each Part: First, differentiate the outer function with respect to the inner function \( u \), and then differentiate the inner function with respect to \( x \).
- Combine Using the Chain Rule: The derivative \( \frac{dy}{dx} \) becomes a product of these two derivatives: \( \frac{dy}{du} \cdot \frac{du}{dx} \).
Inverse Hyperbolic Functions
Inverse hyperbolic functions, like \( \operatorname{coth}^{-1} \), are the inverse of the hyperbolic functions (such as \( \operatorname{coth}\), \( \operatorname{sinh}\), etc.). These functions have practical applications, especially in algebra and calculus, because they can often simplify the handling of particular types of integrals and differential equations.
- Understanding \( \operatorname{coth}^{-1} \): The \( \operatorname{coth}^{-1}(x) \) is defined for \(|x| > 1\). It's crucial to remember restrictions like this when working with such functions.
- Derivative of Inverse Hyperbolic Functions: The derivative of \( \operatorname{coth}^{-1}(x) \) is \( -\frac{1}{1-x^2} \). Always check that \(|x| > 1\), to ensure the derivative is valid.
Calculus Problem-Solving
Calculus problem-solving often requires a structured approach to untangle complex relationships within functions. Here, we illustrate this approach using our given problem.
- Breaking Down Problems: Start by identifying different components of the expression you need to differentiate, such as identifying outer and inner functions, like we did using the chain rule.
- Applying Known Derivatives: Using known derivatives (e.g., of inverse hyperbolic functions) simplifies finding the derivative of more complicated expressions.
- Combining Results: After breaking down the problem and finding necessary derivatives, combine them according to the rules of calculus, such as the chain rule, to get the final result.
- Simplifying Expressions: Once you have combined derivatives, simplify the expression if possible. For example, by factoring or cancelling out common terms.
