Mnemonic for derivative of inverse hyperbolic function

January 15, 2023  302 words 2 mins read  Comments

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.

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  • Damodar Rajbhandari
    Written by Damodar Rajbhandari, a PhD student 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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