Findlim $\displaystyle (3^(2+h)^2)-81/h$.

h->0

*Should be 3 raised to the $\displaystyle (2+h)^2$, not (3) $\displaystyle (2+h)^2$*

How do I solve this? So f(x)=$\displaystyle 3^(x^2)$, $\displaystyle f'= a^x =a^xlna$

I got: $\displaystyle f'(2)=3^(2^2)(ln3)$=81(ln3)

Key says:324(ln3).