How to show $(\nabla df)(X,Y)=\nabla^2_{X,Y}f$?

Question

Why $(\nabla df)(X,Y)=\nabla^2_{X,Y}f$ ?

I only know $df(Y)=Y(f)$, I always see $df$ as an element of $T^*M$,and I think $df=f_idx^i,f_i=df(\partial_i)$.

Besides, whether $(\nabla df)(X,Y)=(\nabla_Xdf)(Y)$?

Thanks.

Answer 1

Let $\omega=df$. Then $(D\omega)(X,Y)=(D_X\omega)(Y)=D_X(\omega(Y))-\omega(D_X(Y))$ So if $\omega_p=0$ (i.e p is a critical point) then $(D\omega)(X,Y)=D_X(\omega(Y))=D_X(D_Y(f))$