I'm told to prove or disprove that $4^{\sqrt{n}}$ grows faster than $\sqrt{4^n}$
As n tends to infinity.
From my Previous years Calculus I know that if I take the derivative of two functions, and one is bigger, than it must grow faster.
$f(n) = 4^{\sqrt{n}}$
$g(n) = \sqrt{4^n}$
if
$f'(n)$ is > $g'(n)$ Then $f(n)$ grows faster than $g(n)$, as $n\to\infty$
My Attempt:
$f'(n) = (\ln(4)\times 4^{\sqrt n)})
/2^{\sqrt n}$
$g'(n) = (\ln(4)\times 4^{n/2})
/2$
But Now I'm not sure if f(n) does grow faster, how would I know that it's first derivative is bigger here, just by plugging in numbers?
Best Answer
Hint: Look at the ratio $\dfrac{4^{\sqrt{n}}}{\sqrt{4^n}}$. This is equal to $$\left(\frac{4}{4^{\sqrt{n}/2}}\right)^{\sqrt{n}}.$$