Answer
$f'(x)= -8(-2x+1)^{3}$
$f''(x)= 48(-2x+1)^{2}$
$f''(x)= -192(2x+1)$
$f^{(4)}(x)=-384$
$f^{(5)}(x)=0$
$f^{(6)}(x)=\ldots=f^{(n)}(x)=0$
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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$f(x)=(-2x+1)^{4}$
$f'(x)=4(-2x+1)^{3}(-2)=-8(-2x+1)^{3}$
$f''(x)=-24(-2x+1)^{2}(-2)=48(-2x+1)^{2}$
$f''(x)=96(-2x+1)(-2)=-192(2x+1)$
$f^{(4)}(x)=-384$
$f^{(5)}(x)=0$
$f^{(6)}(x)=\ldots=f^{(n)}(x)=0$
