Total differential of weighted average

I've stumbled upon this problem looking for something else and now it's bugging me. Suppose you have a weighted average of the form

$y = w_1 * x_1 + w_2 * x_2$

and derive the total derivative, which in my opinion should be

$dy = \frac{\partial y}{\partial w_1} dw_1 + \frac{\partial y}{\partial x_1} dx_1 + \frac{\partial y}{\partial w_2} dw_2 + \frac{\partial y}{x_2} dx_2$

which is equivalent to

$dy = x_1dw_1 + w_1dx_1 + x_2dw_2 + w_2dx_2$.

However, this does not seem to hold when I try it with numeric examples. Where's my mistake?

Thanks!

In time period 1 let $x_1 = 2$, $x_2 = 3$, $w_1 = 0.3$, $w_2 = 0.7$, thus $y = 2.7$. In time period two let $x_1 = 3$, $x_2 = 1$, $w_1 = 0.4$ and $w_2 = 0.6$, thus $y = 1.8$. It follows that $-0.9 = x_1 * 0.1 + w_1 * 1 - x_2 * 0.1 - w_2 * 2$. Using the values from time period 2 the rhs gives $0.6$, using the values from time period 1 it gives $-1.2$. That's what's puzzling me.
• Your opinion is wrong. The right form is $dy = x_1dw_1 + w_1dx_1 + x_2dw_2 + w_2dx_2$ and it will hold all numerical examples. – Satish Ramanathan Dec 16 '17 at 14:20
• I cannot really see a difference in the form you proposed vs. the form I proposed. Consider the following example: In time period 1 let $x_1 = 2$, $x_2 = 3$, $w_1 = 0.3$, $w_2 = 0.7$, thus $y = 2.7$. In time period two let $x_1 = 3$, $x_2 = 1$, $w_1 = 0.4$ and $w_2 = 0.6$, thus $y = 1.8$. It follows that $-0.9 = x_1 * 0.1 + w_1 * 1 - x_2 * 0.1 - w_2 * 2$. Using the values from time period 2 the rhs gives $0.6$, using the values from time period 1 it gives $-1.2$. That's what's puzzling me. – Mr. Zen Dec 16 '17 at 14:37
The formula that you calculate using derivatives gives you the instantaneous change at a certain time. When you calculate with difference of $y$, you get the average change. Also note that $x_1,x_2,w_1,w_2$ might not change linearly, so $d_x1=x_1(t_2)-x_1(t_1)$ is just an approximation. Same for the other variables. You need to get more information about how your variables behave. Are you trying to solve a differential equation?