Mnemonic for derivative of inverse hyperbolic function

January 15, 2023  302 words 2 mins read  Join the Discussion

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 functionInverse 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?

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

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  • Damodar Rajbhandari
    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.
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