Let $M$ be a pseudo-Riemannian manifold.
Let $\Omega=\{(x,v)\in TM: |v|<\epsilon\}$. I want to compute the differential of the map \begin{align*} \Omega &\to TM\\ (x,v) &\mapsto (x,\exp_xv) \end{align*} at $(x,0)$. I know that $d_0(\exp_x)=id$. I tried to write the map in local coordinates: Let $\phi_1$ be a local chart around $\exp_x(v)$ and $\phi_2$ be a local chart around $v$. Then in local coordinates, the map becomes $$(x_1,...,x_n,v_1,...,v_n)\mapsto(x_1,...,x_n,(\phi_1\circ\exp_{\phi_1^{-1}(x_1,...,x_n)} \circ\phi_2^{-1})(v_1,...,v_2))$$ What I am confused is how we can take the partial derivative of $(\phi_1\circ\exp_{\phi_1^{-1}(x_1,...,x_n)} \circ\phi_2^{-1})(v_1,...,v_2))$. Thanks!
Call this map ${\rm Exp}\colon \Omega \to TM$. I suspect this might be a XY problem, so: if your goal is showing that ${\rm d}({\rm Exp})_{(x,0)}$ is non-singular, you do not need to compute all the derivative components -- instead think of a block matrix and compute $${\rm d}({\rm Exp})_{(x,0)} = \begin{pmatrix} {\rm Id} & 0 \\ \ast & {\rm d}(\exp_x)_0 \end{pmatrix},$$by placing the "gradients" of the components in ${\rm Exp}$ in block-rows. Since ${\rm d}(\exp_x)_0 = {\rm Id}$, we already know that this matrix is non-singular, even though you don't know what $\ast$ is.