The derivative of a function like this is computed with the following formula:
[tex] (e^{f(x)})' = e^{f(x)}\cdot f'(x) [/tex]
So, in your case, we have
[tex] f'(x) = e^{x^2+17x} \cdot (2x+17) [/tex]
When you evaluate this at 0, you have
[tex] f'(0) = e^0 \cdot 17 = 1\cdot 17=17 [/tex]
