Calculus 8th Edition

$f^{n}(x)=2^{n}e^{2x}$
$f(x)=e^{2x}$ and $f'(x)=f^{1}(x)=2e^{2x}$ Again differentiate $f'(x)=f^{1}(x)=2e^{2x}$ with rest to x, we get $f''(x)=f^{2}(x)=2^{2}e^{2x}$ Now, take third derivative. $f'''(x)=f^{3}(x)=2^{3}e^{2x}$ Proceeding this process up to nth derivative. Hence, $f^{n}(x)=2^{n}e^{2x}$