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# Derivative of arcsech

### Derivative of sech-1(x)

We use the fact from the definition of the inverse that

$\text{sech}(\text{sech}^{-1} \;x) = x$

and the fact that

$\text{sech}'\, x = -\tanh (x) \text{sech} (x)$

Now take the derivative of both sides (using the chain rule on the left hand side) to get

$-\tanh (\text{sech}^{-1} x)\text{sech}(\text{sech}^{-1}\, x)(\text{sech}^{-1}\, x)' = 1$

or

$-x \, \tanh (\text{sech}^{-1}x)(\text{sech}^{-1} \,x)' = 1 \tag{1}$

We know that

$\cosh^2 x - \sinh^2 x = 1$

Dividing by the $$\cosh^2(x)$$ gives

$1 - \tanh^2 (x) = \text{sech}^2\, x$

or

$\tanh x = \sqrt{1-\text{sech}^2 \,x}$

so that

$\tanh (\text{sech}^{-1}\, x) = \sqrt{1-\text{sech}^{-1} \, x} = \sqrt{1-x^2}$

Finally substituting into equation 1 gives

$-x\sqrt{1-x^2} (\text{sech}^{-1}\, x) = 1$

$\text{sech}^{-1} \, x = \dfrac{-1}{x\sqrt{1-x^2}}$