Answer
$$ f'(x)
=4(x+1)(x^2+2x+3)e^{(x^2+2x+3)^2}.$$
Work Step by Step
Recall that $(e^x)'=e^x$
Recall that $(x^n)'=nx^{n-1}$
Since we have
$$ f(x)= e^{(x^2+2x+3)^2}$$
then the derivative $ f'(x)$, using the chain rule, is given by
$$ f'(x)=e^{(x^2+2x+3)^2}((x^2+2x+3)^2)'=e^{(x^2+2x+3)^2}2(x^2+2x+3)(2x+2)\\
=4(x+1)(x^2+2x+3)e^{(x^2+2x+3)^2}.$$
