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Mathematics LibreTexts

Derivative of arcsech

  • Page ID
    526
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    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

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

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

    Larry Green (Lake Tahoe Community College)