This problem is not complete yet: Testing Notation

This problem is not complete yet: Testing Notation

This problem has been driving me up the wall!
Let $G$ be a matrix group and let $g(t)$ and $h(t)$ be paths in $G$ with
$g(0) = I$, $g^\prime(0) = X$, $h(0) = I$, and $h^\prime(0) = Y$. Prove
that the group commutator $g(t)h(t)g^{-1}(t)h^{-1}(t)$ is equal to $I +
[X, Y]t^2 + \mathcal{O}(t^3)$. (I have heard of this result on two
independent occasions, one of which was from my professor, whom I trust
greatly.)
My problem: My calculation keeps showing that the group commutator's
acceleration vanishes...
To simplify notation, I will write $g$ instead of $g(t)$.
(Before leaving the correct solution, please tell me where my solution is
going wrong..)
First, I take for granted that $(g^{-1})^\prime(0) = -g^\prime(0)$. I now
show that \begin{equation} (g^{-1})^{\prime \prime}(0) = 2(g^\prime)^2(0)
- g^{\prime \prime}(0). \end{equation} $I = g^{-1}g$ implies that $0 =
(g^{-1})^\prime g + g^{-1}g^\prime$, so that \begin{equation} 0 =
(g^{-1})^{\prime \prime} g + (g^{-1})^\prime g^\prime + (g^{-1})^\prime
g^\prime + g^{-1} g^{\prime \prime}. \end{equation} Evaluating at $0$ and
rearranging, we obtain the desired result for $(g^{-1})^{\prime
\prime}(0)$.
Taking the first derivative of the group commutator: \begin{equation}
(ghg^{-1}h^{-1})^\prime = (gh)^\prime (hg)^{-1} + gh \left( (hg)^{-1}
\right)^\prime. \end{equation}
Taking the second derivative and evaluating at $0$, we have \begin{align}
\left\{(ghg^{-1}h^{-1})^{\prime \prime} &= (gh)^{\prime \prime}(hg)^{-1} +
(gh)^\prime \left( (hg)^{-1} \right)^\prime + (gh)^\prime \left( (hg)^{-1}
\right)^\prime + gh \left( (hg)^{-1} \right)^{\prime \prime}\right\}
\bigg|_{t=0} \\ &= x+y. \end{align}