# 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.