Problem
Given $y=\mathrm{arcsech}(e^{-x})$ , find the derivative.
My work
let $x=e^{-x}$ so
$y=\mathrm{arcsech}(x)$
$x=\mathrm{sech}(y)$
I then differentiated implicitly which gave me
$\frac{dy}{dx}=\mathrm{\frac{1}{{-sech}(y){tanh}(y)}}$
I built a right angle triangle with the sides c=x a=$\sqrt{{x^2}{-1}}$ b=1 from the equation $x=\mathrm{sech}(y)$
Using the right angle triangle, I came up with the following ratios
$\mathrm{sech}(y)=\frac{x}{1}$
$\mathrm{tanh}(y)=\frac{\sqrt{{x^2}{-1}}}{{1}}$
Substituting the new equations into the derivative gives me
$\frac{dy}{dx}=\mathrm{\frac{1}{{-x}{\sqrt{{x^2}{-1}}}}}$
Since $x=e^{-x}$, then
$\frac{dy}{dx}=\mathrm{\frac{-1}{{e^{-x}}{\sqrt{{e^{-2x}}{-1}}}}}$
Because this also involves the chain rule, I differentiated
$y=\mathrm{e^{-x}}$
$\frac{dy}{dx}=\mathrm{-e^{-x}}$
So
$\frac{dy}{dx}=\mathrm{\frac{-1}{{e^{-x}}{\sqrt{{e^{-2x}}{-1}}}}}\times\mathrm{-e^{-x}}$
I then concluded that
$\frac{dy}{dx}$ = $\frac{1}{\sqrt{{e^{-2x}}{-1}}}$
The answer states
$\frac{dy}{dx}$ = $\frac{1}{\sqrt{{1-}{e^{-2x}}}}$
Not sure how to fix my error.