$Y_i = \beta X_i + u_i$ , $I^u$ start with minimizing the residuals.

\tilde{\beta}_{i-1} = \xi \min \left\{ \xi (Y_i - \tilde{Y})^2 \right\} = \min \left\{ \xi (Y_i - \beta X_i + u_i)^2 \right\}

Derivating the expression for $\min \left\{ \xi (Y_i - \beta X_i + u_i)^2 \right\}$ , and using the chain rule.

\frac{\partial \left( \xi (Y_i - \beta X_i + u_i)^2 \right)}{\partial (\beta)} = \xi \left\{ \frac{\partial (Y_i - \beta X_i + u_i)^2}{\partial (Y_i - \beta X_i)} \times \frac{\partial (Y_i - \beta X_i + u_i)}{\partial (\beta)} \right\} \quad \Rightarrow \quad \xi (-2(Y_i - \beta X_i) \quad )(0 - X_i) = 0

Removing $-2$ by dividing both sides with $-2$ .

$\xi (Y_i - \beta X_i) \quad )(X_i) = 0$

Multiplying with $X_i$ .

$\xi (Y_i X_i) - \xi (\beta X_i^2) = 0$

Move over $\xi (\beta X_i^2)$ , by addition.

$\xi (Y_i X_i) = \xi (\beta X_i^2)$

Pull $\beta$ out of the expression, and divide by $\xi (X_i^2)$ .

\frac{\xi (Y_i X_i)}{\xi (X_i^2)} = \beta \Rightarrow \beta = \frac{\xi (Y_i X_i)}{\xi (X_i^2)}.