Derivative of Inverse Hyperbolic Coth

Oct 13, 2025 by thesweetdesigns.com 133 views

Assignment 6: Finding Derivatives of Inverse Hyperbolic Functions

In this assignment, we will be finding the derivatives of two functions that involve inverse hyperbolic functions. The inverse hyperbolic functions are a set of functions that are the inverse of the hyperbolic functions. They are used to solve equations that involve hyperbolic functions and are an essential part of calculus.

Derivative of Inverse Hyperbolic Coth

The first function we will be finding the derivative of is:

y = (1-t) \operatorname{coth}^{-1} \sqrt{t}

To find the derivative of this function, we will use the chain rule and the fact that the derivative of the inverse hyperbolic coth function is:

\frac{d}{dx} \operatorname{coth}^{-1} x = \frac{1}{1-x^2}

We will also need to use the fact that the derivative of the square root function is:

\frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}}

Using the chain rule, we can find the derivative of the function as follows:

\frac{dy}{dt} = \frac{d}{dt} (1-t) \operatorname{coth}^{-1} \sqrt{t}

= (1-t) \frac{d}{dt} \operatorname{coth}^{-1} \sqrt{t} + \operatorname{coth}^{-1} \sqrt{t} \frac{d}{dt} (1-t)

= (1-t) \frac{1}{1-(\sqrt{t})^2} \frac{d}{dt} \sqrt{t} + \operatorname{coth}^{-1} \sqrt{t} (-1)

= (1-t) \frac{1}{1-t} \frac{1}{2\sqrt{t}} + \operatorname{coth}^{-1} \sqrt{t} (-1)

= \frac{1}{2\sqrt{t}} + \operatorname{coth}^{-1} \sqrt{t} (-1)

Simplifying the expression, we get:

\frac{dy}{dt} = \frac{1}{2\sqrt{t}} - \operatorname{coth}^{-1} \sqrt{t}

Derivative of Inverse Hyperbolic Coth with t^2

The second function we will be finding the derivative of is:

y = \left(1-t^2\right) \operatorname{coth}^{-1} t

To find the derivative of this function, we will use the chain rule and the fact that the derivative of the inverse hyperbolic coth function is:

\frac{d}{dx} \operatorname{coth}^{-1} x = \frac{1}{1-x^2}

We will also need to use the fact that the derivative of the square function is:

\frac{d}{dx} x^2 = 2x

Using the chain rule, we can find the derivative of the function as follows:

\frac{dy}{dt} = \frac{d}{dt} \left(1-t^2\right) \operatorname{coth}^{-1} t

= \left(1-t^2\right) \frac{d}{dt} \operatorname{coth}^{-1} t + \operatorname{coth}^{-1} t \frac{d}{dt} \left(1-t^2\right)

= \left(1-t^2\right) \frac{1}{1-t^2} \frac{d}{dt} t + \operatorname{coth}^{-1} t (-2t)

= \frac{1}{1-t^2} + \operatorname{coth}^{-1} t (-2t)

Simplifying the expression, we get:

\frac{dy}{dt} = \frac{1}{1-t^2} - 2t \operatorname{coth}^{-1} t

In this assignment, we found the derivatives of two functions that involve inverse hyperbolic functions. The first function was:

y = (1-t) \operatorname{coth}^{-1} \sqrt{t}

The derivative of this function is:

\frac{dy}{dt} = \frac{1}{2\sqrt{t}} - \operatorname{coth}^{-1} \sqrt{t}

The second function was:

y = \left(1-t^2\right) \operatorname{coth}^{-1} t

The derivative of this function is:

\frac{dy}{dt} = \frac{1}{1-t^2} - 2t \operatorname{coth}^{-1} t

These derivatives can be used to solve problems that involve inverse hyperbolic functions and are an essential part of calculus.
Q&A: Finding Derivatives of Inverse Hyperbolic Functions

In our previous article, we found the derivatives of two functions that involve inverse hyperbolic functions. In this article, we will answer some common questions that students often have when it comes to finding derivatives of inverse hyperbolic functions.

Q: What is the derivative of the inverse hyperbolic coth function?

A: The derivative of the inverse hyperbolic coth function is:

\frac{d}{dx} \operatorname{coth}^{-1} x = \frac{1}{1-x^2}

Q: How do I find the derivative of a function that involves the inverse hyperbolic coth function?

A: To find the derivative of a function that involves the inverse hyperbolic coth function, you will need to use the chain rule and the fact that the derivative of the inverse hyperbolic coth function is:

\frac{d}{dx} \operatorname{coth}^{-1} x = \frac{1}{1-x^2}

You will also need to use the fact that the derivative of the square root function is:

\frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}}

Q: What is the derivative of the inverse hyperbolic coth function with t^2?

A: The derivative of the inverse hyperbolic coth function with t^2 is:

\frac{d}{dt} \operatorname{coth}^{-1} t = \frac{1}{1-t^2}

Q: How do I find the derivative of a function that involves the inverse hyperbolic coth function with t^2?

A: To find the derivative of a function that involves the inverse hyperbolic coth function with t^2, you will need to use the chain rule and the fact that the derivative of the inverse hyperbolic coth function with t^2 is:

\frac{d}{dt} \operatorname{coth}^{-1} t = \frac{1}{1-t^2}

You will also need to use the fact that the derivative of the square function is:

\frac{d}{dx} x^2 = 2x

Q: What are some common mistakes to avoid when finding derivatives of inverse hyperbolic functions?

A: Some common mistakes to avoid when finding derivatives of inverse hyperbolic functions include:

Forgetting to use the chain rule
Forgetting to use the fact that the derivative of the inverse hyperbolic coth function is:

\frac{d}{dx} \operatorname{coth}^{-1} x = \frac{1}{1-x^2}

Forgetting to use the fact that the derivative of the square root function is:

\frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}}

In this article, we answered some common questions that students often have when it comes to finding derivatives of inverse hyperbolic functions. We also provided some tips and tricks for avoiding common mistakes when finding derivatives of inverse hyperbolic functions. By following these tips and tricks, you can become more confident and proficient in finding derivatives of inverse hyperbolic functions.

If you are looking for additional resources to help you learn more about finding derivatives of inverse hyperbolic functions, here are a few suggestions:

Khan Academy: Derivatives of Inverse Hyperbolic Functions
MIT OpenCourseWare: Calculus II
Wolfram Alpha: Derivatives of Inverse Hyperbolic Functions

By using these resources, you can get more practice and experience with finding derivatives of inverse hyperbolic functions, and become more confident and proficient in this area of calculus.