#### Answer

$\frac{{e}^{-t}}{\sqrt{( {({e}^{-2t})-1})}}$

#### Work Step by Step

Given that $y$ = $cos^{-1}{{e}^{-t}}$, on substituting ${{e}^{-t}}$ = $u$ and applying chain rule derivative, we get:
$\frac{dy}{dt}$ = $\frac{{dcos^{-1}{u}}}{du}$ $\times$ $\frac{du}{dt}$
or $\frac{dy}{dt}$ = $-\frac{1}{\sqrt {(1-{u^2})}}$ $\times$ $\frac{du}{dt}$
so $\frac{dy}{dt}$ = $-\frac{1}{\sqrt {({1-{e}^{-2t}}})}$ $\times$ $\frac{d {{e}^{-t}}}{dt}$
or $\frac{dy}{dt}$ = $\frac{1}{\sqrt{( {({e}^{-2t})-1})}}$ $\times$ ${e}^{-t}$
The final answer is: = $\frac{{e}^{-t}}{\sqrt{( {({e}^{-2t})-1})}}$