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Derivative of Inverse Hyperbolic Trigonometry: coth^-1(x)

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First published at 17:43 UTC on May 25th, 2020.

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In this video I go over the derivative of inverse hyperbolic cotangent or coth^-1(x) and show that it is equal to 1/(1-x^2).

View video notes on the Hive b…

In this video I go over the derivative of inverse hyperbolic cotangent or coth^-1(x) and show that it is equal to 1/(1-x^2).

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