Simple proof verification

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I have the following equations:

\begin{equation}\boldsymbol{\beta}_i(x_1,\ldots,x_i) = \Gamma_i \int_{0}^{x_i}\boldsymbol{\varphi}_i(x_1,\ldots,x_{i-1},\chi)\, d \chi + \boldsymbol{\epsilon}_i(x_i)\end{equation}

\begin{equation}F_i(x_1,\ldots,x_i) = \Gamma_i \sum_{k=1}^{i-1}\frac{\partial }{\partial x_{k}}\left ( \int_{0}^{x_{i}} \boldsymbol{\varphi}_i(x_1,\ldots,x_{i-1},\chi) \,d\chi\right )\boldsymbol{\varphi}_k(x_1,\ldots,x_k)^T + \frac{\partial \boldsymbol{\epsilon}_i}{\partial x_i}\boldsymbol{\varphi}_i(x_1,\ldots,x_i)^T\end{equation}

where $\Gamma_i$ is a constant matrix.

Now the following equality should hold according to my book: \begin{equation}\sum_{k=1}^{i}\frac{\partial \boldsymbol{\beta}_i}{\partial x_k} \boldsymbol{\varphi}_k(x_1,\ldots,x_k)^T = 2\Gamma_i\boldsymbol{\varphi}_i\boldsymbol{\varphi}_i^T + F_i + F_i^T\end{equation}

So I try to check whether this is correct: \begin{align} \sum_{k=1}^{i}\frac{\partial \boldsymbol{\beta}_i}{\partial x_k} \boldsymbol{\varphi}_k(x_1,\ldots,x_k)^T &= \sum_{k=1}^{i-1}\frac{\partial }{\partial x_k} \left \{ \Gamma_i \int_{0}^{x_i}\boldsymbol{\varphi}_i(x_1,\ldots,x_{i-1},\chi)d\chi \right \}\boldsymbol{\varphi}_k(x_1,\ldots,x_k)^T + \frac{\partial }{\partial x_i} \left \{ \Gamma_i \int_{0}^{x_i}\boldsymbol{\varphi}_i(x_1,\ldots,x_{i-1},\chi)d\chi + \boldsymbol{\epsilon}_i(x_i)\right \}\boldsymbol{\varphi}_i(x_1,\ldots,x_i)^T \\ &= \sum_{k=1}^{i-1}\frac{\partial }{\partial x_k} \left \{ \Gamma_i \int_{0}^{x_i}\boldsymbol{\varphi}_i(x_1,\ldots,x_{i-1},\chi)d\chi \right \}\boldsymbol{\varphi}_k(x_1,\ldots,x_k)^T + \Gamma_i \boldsymbol{\varphi}_i(x_1,\ldots,x_i)\boldsymbol{\varphi}_i(x_1,\ldots,x_i)^T+\frac{\partial \boldsymbol{\epsilon}_i(x_i)}{\partial x_i}\boldsymbol{\varphi}_i(x_1,\ldots,x_i)^T \\ &= \Gamma_i \boldsymbol{\varphi}_i(x_1,\ldots,x_i)\boldsymbol{\varphi}_i(x_1,\ldots,x_i)^T +F_i \end{align}

Now I don't know how to continue.