Answer
$f'(x)=-9x^{2}+4$
$f''(x)=-18x$
$f'''(x)=-18$
$f^{(4)}(x)=0$
$f^{(5)}(x)\ldots=f^{(n)}(x)=0$
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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$f(x)=-3x^{3}+4x$
$f'(x)=-9x^{2}+4$
$f''(x)=-18x$
$f'''(x)=-18$
$f^{(4)}(x)=0$
$f^{(5)}(x)\ldots=f^{(n)}(x)=0$
