Mnemonic for derivative of inverse hyperbolic function
Today, I remember a mnemonic, and thought I should share it with you. I hope it will help you to remember the derivative of inverse hyperbolic function. When I was taking Calculus course in my high school, my teacher told us to memorize for the next class; the derivative of inverse trigonometry function and inverse hyperbolic function. First, I start memorizing the derivative of all the inverse trignometry functions. As soon as I finished memorizing for the second ones, I kept on forgetting the first ones. I realised it quickly that they are seemingly similar expression. So, I made the mnemonic.
Before I tell you the mnemonic, please see the below table of formulae. You will notice the same thing I realised.
| Inverse trigonometry function | Inverse hyperbolic function |
|---|---|
| $\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1 - x^{2}}}$ | $\frac{d}{dx} \sinh^{-1}(x) = \frac{1}{\sqrt{1 + x^{2}}}$ |
| $\frac{d}{dx} \cos^{-1}(x) = \frac{-1}{\sqrt{1 - x^{2}}}$ | $\frac{d}{dx} \cosh^{-1}(x) = \frac{1}{x^{2} - 1}$ |
| $\frac{d}{dx} \tan^{-1}(x) = \frac{1}{1 + x^{2}}$ | $\frac{d}{dx} \tanh^{-1}(x) = \frac{1}{1 - x^{2}}$ |
| $\frac{d}{dx} \cot^{-1}(x) = \frac{-1}{x^{2} + 1}$ | $\frac{d}{dx} \coth^{-1}(x) = \frac{-1}{x^{2} - 1}$ |
| $\frac{d}{dx} \sec^{-1}(x) = \frac{1}{x \sqrt{x^{2} - 1}}$ | $\frac{d}{dx} \text{sech}^{-1}(x) = \frac{-1}{x \sqrt{1 - x^{2}}}$ |
| $\frac{d}{dx} \text{cosec}^{-1}(x) = \frac{-1}{x\sqrt{x^{2} - 1}}$ | $\frac{d}{dx} \text{cosech}^{-1}(x) = \frac{-1}{x \sqrt{x^{2} + 1}}$ |
Here the mnemonic goes like this:
- First of all, remember the derivative of the inverse trigonometry function exactly as I written in the first column.
- In order to memorize the formulae for the derivative of the inverse hyperbolic function, use the below rule:
- For $\cosh^{-1}$ and $\text{sech}^{-1}$, multiply by minus in the numerator and inside the square root in the denominator of $\cosh^{-1}$ and $\text{sech}^{-1}$ respectively i.e. $\frac{\text{up}}{\sqrt{\text{down}}} \to \frac{\text{- up}}{\sqrt{\text{- down}}}$.
- For others, change the sign of second term in the denominator. i.e. $\pm \to \mp$
I hope this helps you to remember the formulae with less effort.
Any feedback?
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Written by Damodar Rajbhandari, a PhD candidate in Theoretical Physics at the Jagiellonian University, Poland, working in the Causal Dynamical Triangulations group at the Institute of Theoretical Physics.
