calculus - Proving that a function grows faster than another

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?

Answer

Hint: Look at the ratio $\dfrac{4^{\sqrt{n}}}{\sqrt{4^n}}$. This is equal to

$$\left(\frac{4}{4^{\sqrt{n}/2}}\right)^{\sqrt{n}}.$$