calculus – Time derivative of a gradient

There is this equation $$ ddot{X}_t + (e^{alpha_t}-dot{alpha_t})dot{X}_t + e^{2alpha_t+beta_t}(nabla^2h(X_t + e^{-alpha_t}dot{X}_t))^{-1}nabla f(X_t)=0 $$
Now, the paper says that the equation can be written in the following form $$ frac{d}{dt}nabla h(X_t + e^{-alpha_t}dot{X}_t)) = -e^{alpha_t+beta_t}nabla f(X_t)$$

If I want the original equation back then what will the time derivative of the gradient $nabla h(X_t + e^{-alpha_t}dot{X}_t)$.

If I take the time derivative to be $nabla^2 h(X_t + e^{-alpha_t}dot{X}_t)(dot{X}_t+ddot{X}_t e^{-alpha_t} -alpha dot{X}_t e^{-alpha_t})$ then I get the above 2nd degree differential equation but the main question is how come the time derivative of a gradient becomes a hessian?
Also, the function $h:R^nrightarrow R$ and X $subset R^n$. Also, here $nabla^2$ is hessian as said in the paper not Laplacian operator. Also, $t$ in $X_t$ and $alpha_t$ means they are a function of time.