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[{{:r.regressionline3.png}}]

\begin{eqnarray*}
\sum{(Y_i - \hat{Y_i})^2} & = & \sum{(Y_i - (a + bX_i))^2}  \;\;\; \because \hat{Y_i} = a + bX_i \\
& = & \text{SSE or SS.residual} \;\;\; \text{(and this should be the least value.)}
\end{eqnarray*}

<WRAP box>
\begin{eqnarray*}
\text{for a (constant)} \\ 
\\
\dfrac{\text{d}}{\text{dv}} \sum{(Y_i - (a + bX_i))^2} & = & \sum \dfrac{\text{d}}{\text{dv}} {(Y_i - (a + bX_i))^2} \\ 
& = & \sum{2 (Y_i - (a + bX_i))} * (-1) \;\;\;\; \\
& \because & \dfrac{\text{d}}{\text{dv for a}} (Y_i - (a+bX_i)) = -1 \\
& = & -2 \sum{(Y_i - (a + bX_i))} \\ 
\\
\text{in order to have the least value, the above should be zero} \\ 
\\
-2 \sum{(Y_i - (a + bX_i))} & = &  0 \\
\sum{(Y_i - (a + bX_i))} & = & 0 \\ 
\sum{Y_i} - \sum{a} - b \sum{X_i} & = & 0 \\
\sum{Y_i} - n*{a} - b \sum{X_i} & = & 0 \\
n*{a} & = & \sum{Y_i} - b \sum{X_i} \\
a & = & \dfrac{\sum{Y_i}}{n} - b \dfrac{\sum{X_i}}{n} \\
a & = & \overline{Y} - b \overline{X} \\
\end{eqnarray*} 
</WRAP>

<WRAP box>
\begin{eqnarray*}
\text{for b, (coefficient)} \\ 
\\
\dfrac{\text{d}}{\text{dv}} \sum{(Y_i - (a + bX_i))^2}  & = & \sum \dfrac{\text{d}}{\text{dv}} {(Y_i - (a + bX_i))^2} \\ 
& = & \sum{2 (Y_i - (a + bX_i))} * (-X_i) \;\;\;\; \\
& \because & \dfrac{\text{d}}{\text{dv for b}} (Y_i - (a+bX_i)) = -X_i \\
& = & -2 \sum{X_i (Y_i - (a + bX_i))} \\
\\
\text{in order to have the least value, the above should be zero} \\ 
\\
-2 \sum{X_i (Y_i - (a + bX_i))} & = & 0 \\
\sum{X_i (Y_i - (a + bX_i))} & = & 0 \\ 
\sum{X_i (Y_i - ((\overline{Y} - b \overline{X}) + bX_i))} & = & 0 \\ 
\sum{X_i ((Y_i - \overline{Y}) - b (X_i - \overline{X})) } & = & 0 \\ 
\sum{X_i (Y_i - \overline{Y})} - \sum{b X_i (X_i - \overline{X}) } & = & 0 \\ 
\sum{X_i (Y_i - \overline{Y})} & = &  b \sum{X_i (X_i - \overline{X})} \\ 
b & = & \dfrac{\sum{X_i (Y_i - \overline{Y})}}{\sum{X_i (X_i - \overline{X})}} \\
b & = & \dfrac{\sum{(Y_i - \overline{Y})}}{\sum{(X_i - \overline{X})}} \\
b & = & \dfrac{ \sum{(Y_i - \overline{Y})(X_i - \overline{X})} } {\sum{(X_i - \overline{X})(X_i - \overline{X})}} \\
b & = & \dfrac{ \text{SP} } {\text{SS}_\text{x}} \\
\end{eqnarray*} 
</WRAP>