In a concrete manifold of $\mathbb{R}^3$ (namely, the $2$-sphere of radius $a$), I obtained he following exponential map $$ \exp_pX=p\cos(\frac{\parallel X \parallel}{a}) + \frac{X}{\parallel X \parallel}a\sin(\frac{\parallel X \parallel}{a}) $$ And now I want to prove that its differential at zero is the identity.
My attempt:
I have
$$d_0\exp_p:T_0(T_pM)=T_pM\rightarrow T_{exp_p0=p}M$$
This is a linear map. If I prove that the associated matrix is the identity, I'm done. Now, a basis for $T_pM$ is $\{ \partial_\phi, \partial_\theta \}$. $T_pM$ is natural homeomorphic to $\mathbb{R}^2$ with a global chart $(x^1,x^2)\overset{\Psi}{\rightarrow} x^1\partial_\phi + x^2\partial_\theta$. Now, the entries of the Jacobian matrix are $\frac{\partial(\exp_p \circ \Psi)_i}{\partial x^j}$. I have $$ \frac{\partial(\exp_p \circ \Psi)}{\partial x^1}= \frac{\partial}{\partial x^1} \Big(p\cos(\frac{\parallel (x^1,x^2) \parallel}{a}) + \frac{(x^1,x^2)}{\parallel (x^1,x^2) \parallel}a\sin(\frac{\parallel (x^1,x^2) \parallel}{a})\Big) $$
I can certainly derive the second term. But the first one does not make sense: how do I deal with $p$? I cannot write it in the coordinates of the chart... Also, the derivative of the second term gives something very far from $(1,0)$ (which we would expect since this is the first row of the Jacobian matrix).
Any hints?
(Context: Manifolds in $\mathbb{R}^n$ (i.e not in the context of general smooth manifolds)).
The way I usually think about the exponential map is in terms of radial lines through the origin. That is instead of trying to calculate the Jacobian in coordinate charts what one can do is consider a fixed $X \in T_{p}M$ and consider the curve $\gamma(t) = tX \in T_{p}M$. Then to show that the differential of $\exp_{p}: T_{p}M \to S^{2}$ is the identity it suffices to show that$\frac{d}{dt}\bigg\vert_{t = 0} \exp_{p}(\gamma(t)) = X$. To see this think about what the chain rule tells you.