Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

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

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

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

Contributors