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