Answer
$f'(x)=-e^{-x}$
$f'(x)=e^{-x}$
$f''(x)-e^{-x}$
$f^{(4)}(x)=e^{-x}$...
$f^{(n)}(x)=(-1)^{n}e^{-x}$
Work Step by Step
$f'''(x)=\displaystyle \frac{d}{dx}\left[f''(x)\right]$
$f^{(4)}(x)=\displaystyle \frac{d}{dx}\left[f'''(x)\right]$
$\displaystyle \quad f^{(5)}(x)=\frac{d}{dx}\left[f^{(4)}(x)\right]$
and so on, assuming all these derivatives exist.
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$f(x)=e^{-x}$
$f'(x)=e^{-x}(-1)=-e^{-x}$
$f'(x)=-e^{-x}(-1)=e^{-x}$
$f''(x)=e^{-x}(-1)=-e^{-x}$
$f^{(4)}(x)=-e^{-x}(-1)=e^{-x}$
$\cdots$
we observe that the sign is positive for even n, and the factor $e^{-x}$ appears in each derivative.
$f^{(n)}(x)=(-1)^{n}e^{-x}$
