Derivative of arcsech
- Page ID
- 526
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)