Derivative of a composite quotient using chain and quotient rule

Let $u = e^{x^2}$ and $v = (x+1)^4$. By the chain rule, $u' = e^{x^2}\cdot 2x$, and by the power/chain rule, $v' = 4(x+1)^3$. By the quotient rule, $f'(x) = \dfrac{u'v - uv'}{v^2} = \dfrac{2x e^{x^2}(x+1)^4 - e^{x^2}\cdot 4(x+1)^3}{(x+1)^8}$.

Now substitute $x=0$: $u(0) = e^0 = 1$, $u'(0) = 2(0)e^0 = 0$, $v(0) = 1^4 = 1$, and $v'(0) = 4(1)^3 = 4$. So $f'(0) = \dfrac{u'(0)v(0) - u(0)v'(0)}{v(0)^2} = \dfrac{0\cdot 1 - 1\cdot 4}{1^2} = -4$.

The outer function is the exponential function, written as e to the something. The inner function was x squared, which gets plugged into the outer exponential function. Recognizing this split was necessary to apply the chain rule and get the derivative two x times e to x squared.