The derivative of the cost/loss function

Computing $\partial C(\beta) / \partial \beta_0$ and $\partial C(\beta) / \partial \beta_1$ we can show that the gradient can be written as $$\nabla_{\beta} C(\beta) = (\partial C(\beta) / \partial \beta_0, \partial C(\beta) / \partial \beta_1)^T = 2\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\ \sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\ \end{bmatrix} = 2X^T(X\beta - \mathbf{y}),$$ where $X$ is the design matrix defined above.

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