Answer
$f'(x)=-2e^{-x+3}$
$f''(x)=2e^{-x+3}$
$f''(x)=-2e^{-x+3}$
$f^{(4)}(x)=2e^{-x+3}$
$\cdots$
$f^{(n)}(x)=(-1)^{n}2e^{-x+3}$
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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$f(x)=2e^{-x+3}$
$f'(x)=2e^{-x+3}(-1)=-2e^{-x+3}$
$f''(x)=-2e^{-x+3}(-1)=2e^{-x+3}$
$f''(x)=2e^{-x+3}(-1)=-2e^{3x-1}$
$f^{(4)}(x)=-2e^{-x+3}(-1)=2e^{3x-1}$
$\cdots$
$f^{(n)}(x)=(-1)^{n}2e^{-x+3}$
