Derivative of arcsech

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

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

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

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

We know that

\[ \cosh^2 x - \sinh^2 x = 1\]

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

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

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

so that

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

Finally substituting into equation 1 gives

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

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