(Requires some Calculus)Consider the sample regression function . $Y _ { i } = \hat { \beta } _ { 0 } + \hat { \beta } _ { 1 } X _ { 1 i } + \hat { \beta } _ { 2 } X _ { 2 i }$ Take the total derivative. Next show that the partial derivative $\frac { \Delta Y _ { i } } { \Delta X _ { 1 i } }$ is obtained by holding $X _ { 2 i }$ constant, or controlling for $X _ { 2 i }$

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The Answer of (Requires some Calculus)Consider the sample regression function . $Y _ { i } = \hat { \beta } _ { 0 } + \hat { \beta } _ { 1 } X _ { 1 i } + \hat { \beta } _ { 2 } X _ { 2 i }$ Take the total derivative. Next show that the partial derivative $\frac { \Delta Y _ { i } } { \Delta X _ { 1 i } }$ is obtained by holding $X _ { 2 i }$ constant, or controlling for $X _ { 2 i }$

(Requires Some Calculus)Consider the Sample Regression Function Yi=β^0+β^1X1i+β^2X2iY _ { i } = \hat { \beta } _ { 0 } + \hat { \beta } _ { 1 } X _ { 1 i } + \hat { \beta } _ { 2 } X _ { 2 i }Yi​=β^​0​+β^​1​X1i​+β^​2​X2i​

(Requires Some Calculus)Consider the Sample Regression Function $Y _ { i } = \hat { \beta } _ { 0 } + \hat { \beta } _ { 1 } X _ { 1 i } + \hat { \beta } _ { 2 } X _ { 2 i }$