Let $X=\sum_{i=1}^nA_i\frac{\partial}{\partial x_i}$ be a vector field in $\mathbb{R^n}$ which has a restriction $\tilde{X}$ to the sphere $S^{n-1}$. Let $c:I\rightarrow S^{n-1}$ ,$c(t)=(c_1(t),...,c_n(t))$ be a smooth curve. Define $B_i(t)=\frac{d(A_i(c(t))}{dt}$, $i=1,2,...,n$. Show that $\frac{D\tilde{X}}{dt}=\sum_{i=1}^n (B_i(t)-\langle B(t),c(t)\rangle c_i(t))\frac{\partial}{\partial x_i}$.
So the formula is $\frac{D\tilde{X}}{dt}=\sum_j\frac{dA_j(c(t))}{dt}\frac{\partial}{\partial x_i}+\sum_{i,j}\frac{dc_i(t)}{dt}A_j(c(t)\nabla_{\frac{\partial}{\partial x_i}}\frac{\partial}{\partial x_j}$. So now I am really confused with the terms $\nabla_{\frac{\partial}{\partial x_i}}\frac{\partial}{\partial x_j}$. What are they? How do we define them?
Hint :
(1) Note that if $X=\sum_{i=1}^n A_i\frac{\partial }{\partial x_i}$, then $\frac{\partial }{\partial x_i}$ is not coordinate vector field on sphere
(2) If $N\subset M$ and $N$ has induced metric from $M$, then covariant derivative on $N$ is a projection of convariant derivative on $M$
If $M=\mathbb{R}^n$, then covariant derivative on $M$ is a just differentiation Hence we can complete the proof