If f(x)=e^x g(x), g(0)=2, g'(0)=1 then f'(0) is equal to :

$f(x) = e^xg(x)~~~g(0)=2~~~g'(0)=1$

so  $f(0) = e^0(g(0))$

f(0) = 2

differentiating f(x) wrt x

using product rule

so  $f'(x) = \frac{d}{dx} (e^xg(x))$

so  $f'(x) = e^xg'(x) + e^xg(x)$

so f'(0) = e⁰g'(0) + e⁰g(0)

= 2 + 1 = 3