The Definition of the Second Fundamental Form Let $r:M\rightarrow{\mathbb{R}^{n+1}}$ be an isometric immersion and $M$ is an $n$-dimensional Riemannian Manifold. That is to say, $M$ is the hypersurface in $\mathbb{{R}^{n+1}}$.

Then we can introduce a normal vector field: $N:M\rightarrow{T\mathbb{{R}^{n+1}}}=\mathbb{R}^{n+1}\times{\mathbb{R}^{n+1}}$ satisfies $N_p\in{T_{r(p)}\mathbb{R}^{n+1}}=\{r(p)\}\times\mathbb{R}^{n+1}$. So we will have $T_{r(p)}\mathbb{R}^{n+1}=r_*(T_pM)\oplus{span\{N_p\}}$.
Before we talk about this problem, we look at the connection on $\mathbb{R}^{n+1}$. Let $\bigtriangledown$ be its connection and $X=\sum_{i=1}^{n+1}x_ie_i$, $Y=\sum_{i=1}^{n+1}y_ie_i$. So $\bigtriangledown_XY=\sum_{i=1}^{n+1}\sum_{j=1}^{n+1}x_je_j(y_i)e_i$.

When I read book, I find two different definitions of The Second Fundamental Form. I want to verify that they are the same.

Let $\bar{n}$ denote the Guass Map which is actually: $n=\pi_2\circ{N}:M\rightarrow{\mathbb{R}^{n+1}}$.
The first definition is:

$II|_U=\sum_{i=1}^{n}\sum_{j=1}^{n}<\bigtriangledown_{\frac{\partial{r}}{\partial{x_i}}}N,\frac{\partial{r}}{\partial{x_j}}>dx_i\otimes{dx_j}$

The second definition is:

$II|_U=\sum_{i=1}^{n}\sum_{j=1}^{n}<\frac{\partial{n}}{\partial{x_i}},\frac{\partial{r}}{\partial{x_j}}>dx_i\otimes{dx_j}$


So I think they are the same. Then I try to prove it. But I am failed.

I want to prove $\bigtriangledown_{\frac{\partial{r}}{\partial{x_i}}}N=\frac{\partial{n}}{\partial{x_i}}$.

Proof. Let $\frac{\partial{r}}{\partial{x_i}}=\sum_{k=1}^{n+1}\frac{\partial{r_k}}{\partial{x_i}}e_k$ and $N=\sum_{k=1}^{n+1}n_ke_k$.
Then I have no idea.


How to prove??
 A: Hint:
Consider that if $\langle\partial_j,n\rangle=0$ then also $\langle D_{\partial_i}\partial_j,n\rangle+\langle\partial_j,D_{\partial_i}n\rangle=0$. Use it to correct your guessing. 
One is taking $\partial_i=\frac{\partial}{\partial x^i}$ for the coordinated tangent vectors. 
Update:
Very probable you are using the standard covariant derivative $\nabla$ of ${\Bbb{R}}^{n+1}$.
In this case one computes for $\nabla_XY=[JY]X$ for vector fileds $X,Y$ where $[JY]$ means the 
Jacobian  of the field $Y$.
The covariant derivative and the pairing satisfy:
$$
\frac{\partial}{\partial x^i}\langle X,Y\rangle=
\langle \nabla_{\frac{\partial}{\partial x^i}}X,Y\rangle
+
\langle X,\nabla_{\frac{\partial}{\partial x^i}}Y\rangle
.$$
When applied to $N$ and $\frac{\partial}{\partial x^j}$
which $\langle N,\frac{\partial}{\partial x^i}\rangle=0$ then
\begin{eqnarray*}
\frac{\partial}{\partial x^i}\langle N,\frac{\partial}{\partial x^j}\rangle&=&0\\
\langle \nabla_{\frac{\partial}{\partial x^i}}N,\frac{\partial}{\partial x^j}\rangle+\langle N,\nabla_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}\rangle&=&0\\
\langle \nabla_{\frac{\partial}{\partial x^i}}N,\frac{\partial}{\partial x^j}\rangle&=&-
\langle N,\nabla_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}\rangle
\end{eqnarray*}
So 
$$II|_U=\sum_{ij}
\langle\nabla_{\frac{\partial{}}{\partial{x^i}}}N,\frac{\partial{}}{\partial{x^j}}
\rangle dx^i\otimes{dx^j},
$$
is also
$$\qquad
=-\sum_{ij}
\langle N,\nabla_{\frac{\partial{}}{\partial{x^i}}}\frac{\partial{}}{\partial{x^j}}
\rangle dx^i\otimes{dx^j}.
$$
