Let $M$ be a Riemannian manifold. Let us look at the Riemannian exponential function $\exp_x: T_x M \supset \mathcal{D} \longrightarrow M$.

The derivative of the exponential map can be expressed in Terms of Jacobi Fields. Is there any slick way to express the *second* derivative 
$$ \nabla d \exp_x|_X: T_x M \times T_x M \longrightarrow T_{\exp_x(X)}M$$
in terms of integrals over curvature or Jacobi Fields? What about its Taylor expansion in $X \in T_x M$ about $0$?

(Just to clarify what I mean with the above expression: We can pullback the tangent bundle $TM$ and the Levi-Civita connection to $T_x M$. Look at the vector bundle $T^*T_x M \otimes \exp_x^* TM \rightarrow T_x M$. The first factor is naturally isomorphic to $T^*_x M$ and is just flat; for the second one, we use the pullback of the Levi-Civita Connection. Hence we have a vector bundle with connection, and $d \exp_x$ is a section of it. Hence we can form $\nabla$ of it, which will be a section of $T^*_x M \times T^*_x M \otimes \exp_x^* TM$; that is the quantity I am interested in.) 

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**Edit:** Let $c(t, \varepsilon) := \exp_x(t(X+\varepsilon Y))$. Then
$$ (d \exp_x|_{t(X+\varepsilon Y)}) \cdot t Z = J(t, \varepsilon),$$
where $J(t, \varepsilon)$ is the Jacobi field along $c(t, \varepsilon)$ with $J(0, \varepsilon) = 0$ and $\frac{\nabla}{\partial t} J(0, \varepsilon) = Z$. Now
$$(\nabla d \exp_x|_{tX})[tZ, tZ] = \frac{\nabla}{\partial \varepsilon} J(t, 0) =: K(t)$$

Now Deane Zang claims in the comments that $K(t)$ fulfills a second order equation obtained by differentiating the equation for $J(t, \varepsilon)$,
$$\left(\frac{\nabla}{\partial t}\right)^2 J(t, \varepsilon) = R\left(\frac{\partial}{\partial t} c(t, \varepsilon), J(t, \varepsilon)\right)\frac{\partial}{\partial t} c(t, \varepsilon).$$
However, when I differentiate this (with respect to $\varepsilon$, I suppose?), I get all kinds of terms, $J(t, 0)$ and its derivative, derivatives of the curvature tensor, and especially second derivatives of $\exp$ in direction $X$ and $Z$. So this seems far from obtaining a good equation for $K$.