Answer
$f^{n}(x)=2^{n}e^{2x}$
Work Step by Step
$f(x)=e^{2x}$
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.
$f^{n}(x)=2^{n}e^{2x}$
Hence, the sequence of nth derivative follows powers of two.
