# 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 an inverse hyperbolic function. When I was taking a Calculus course in 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 the second one, I kept on forgetting the first one. I realised quickly that they were seemingly similar expressions. So, I made the mnemonic.

Before I tell you the mnemonic, please see the below table of formulae. You may 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 wrote 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 the second term in the denominator. i.e. $\pm \to \mp$

I hope this helps you to remember the formulae with less effort.

Any feedback?If you guys have some questions, comments, or suggestions then, please don't hesitate to shot me an email at [firstname][AT]physicslog.com or comment below.

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