Finding slope of a tangent to a function

Hi guys,

Another question I'm confused with. Question is:

Find the slope of the tangent to the function f(x)=2^x^2+3x when x=3.

First I start with the chain rule:

f(x)=2^(x^2+3x)

f'(x)=2^(x^2+3x) ln2 (2x+3)

Then I input the value of x=3 into the equation.

f'(3)=2^(3^2+3(3)) ln2 (2(3)+3)

f'(3)=2^(3^2+3(3)) ln2 (9)

This is where I get stuck, if I complete the function I am left with:

f'(3)=2^18ln2 (9)

f'(3)= 1,635,229.37. Obviously this is not the correct answer.

I'm stuck, again.