Chapter 6 - Inverse Functions - 6.7 Hyperbolic Functions - 6.7 Exercises - Page 490: 44

$\dfrac{1}{\sqrt{1-e^{-2x}}}$

Given: $\DeclareMathOperator{\sech}{sech}$$y=\sech^{-1}(e^{-x})$ Now, $y'=\dfrac{d}{dx}\sech^{-1}(e^{-x})$ Apply chain rule, we have $y'=-\dfrac{1}{{e^{-x}}\sqrt{1-({e^{-x}})^2}}(-e^{-x})$ or, $y'=\dfrac{e^{-x}}{e^{-x}\sqrt{1-e^{-2x}}}$ Thus, $y'=\dfrac{1}{\sqrt{1-e^{-2x}}}$