1

Can anyone explain how to partially differentiate the equation $\alpha_1 - \beta(z - x_1) = \alpha_2 - \beta(x_2 - z)$ with respect to $\alpha$ and $x_1$, where $\beta' > 0$, $\beta'' > 0$, $\beta(0) = 0$. I have the following answer: $\frac{\partial z}{\partial \alpha_1} = \frac{1}{\beta'(z - x_1) + \beta'(x_2 - z)}$,

$\frac{\partial z}{\partial x_1} = \frac{\beta'(z - x_1)}{\beta'(z-x_1) +\beta'(x_2 - z)}$

Similarly,

$\frac{\partial z}{\partial \alpha_2} = \frac{1}{\beta'(z - x_1) + \beta'(x_2 - z)}$,

$\frac{\partial z}{\partial x_2} = \frac{\beta'(z - x_2)}{\beta'(z-x_1) +\beta'(x_2 - z)}$

I don't get how they come to this answer. I have $z = \frac{\alpha_1 - \alpha_2}{2\beta} + \frac{x_1 + x_2}{2}$, but then how do partial derivative to come up with those answers?

EditPiAf
  • 20,898
JoeDi
  • 27

1 Answers1

1

The symbol $\beta$ denotes a function, not a constant. Therefore, if $x_{1,2}$ don't depend on $\alpha_1$, partial differentiation of $\alpha_1 - \beta(z-x_1) = \alpha_2 - \beta(x_2-z)$ w.r.t. $\alpha_1$ gives $$ 1 - \frac{\partial z}{\partial \alpha_1} \beta'(z-x_1) = 0 + \frac{\partial z}{\partial \alpha_1} \beta'(x_2-z)\, , $$ where $\beta'$ is the derivative of $\beta$. Hence, $$ \frac{\partial z}{\partial \alpha_1} = \frac{1}{\beta'(z-x_1) + \beta'(x_2-z)}\, , $$ and so on and so forth.

EditPiAf
  • 20,898