Suppose that $\epsilon > 0$, $\gamma: (-\epsilon,\epsilon) \to \text{End}(\mathbb{R}^n)$ is a differentiable curve of matrices such that $\text{det}(\gamma(t)) = 1$ for all $t\in(-\epsilon,\epsilon)$ and $\gamma(0) = I$. Show that $(D\gamma)_0$ has trace zero.
I know that $\gamma(0+h) = \gamma(0)+(D\gamma)_0(h)+\epsilon(h)$ however, I don't know how to calculate the derivative since I don't know the function $\gamma$ explicitly and I am not really sure if there are any properties of the trace that could help me.
Recall that the characteristic polynomial $p_A(t) = \det(tI-A) = t^n -\text{tr}(A)t^{n-1}+\dots+(-1)^n\det A$. From this you should be able to get that $$\frac d{ds}\Big|_{s=0} \det (I+sA) = \text{tr}(A).$$ (Of course, you can get it explicitly by expanding the determinant, too.)