0
$\begingroup$

Say I have the following equation:

$$z = x + y$$

Then we can say that:

$$ \Delta z = \Delta x + \Delta y = (x_1 - x_0) + (y_1 - y_0)$$

Now, say that we have:

$$ z = xy$$

How do I decompose this? It is clearly not the case that:

$$ \Delta z = y_0\Delta x + x_0\Delta y$$

So how do I apportion where the changes are coming from?


For instance, say that $x_0 = 2, x_1 = 4$ and $y_0 = 6, y_1 = 8$.

Then, $z_0 = 12, z_1 = 32$.

How do I apportion how much of the change in $z$ of 20 is coming from $x$ vs $y$?

$\endgroup$
2
  • 1
    $\begingroup$ $\Delta z=z-z_0=xy-x_0y_0=xy-xy_0+xy_0-x_0y_0=x\Delta y+y_0\Delta x=x_0\Delta y+y_0\Delta x+...$ $\endgroup$
    – A.Γ.
    Mar 1, 2020 at 18:08
  • 1
    $\begingroup$ For reference, "Increment Theorem for Functions of Two Variables" $\endgroup$ Mar 1, 2020 at 18:13

1 Answer 1

1
$\begingroup$

Assuming $f$ is differentiable at $(x,y)=(x_0, y_0)$, $$\Delta z = f_x(x_0,y_0)\Delta x + f_y(x_0,y_0)\Delta + \epsilon_1\Delta x + \epsilon_2\Delta y$$ where

$\epsilon_1 = f_x(c_x,y_0) -f_x(x_0,y_0)$,

$\epsilon_2 = f_y(x+\Delta x,d) -f_y(x_0,y_0)$, you get $c_x$ and $d$ by MVT.

For small changes, you can ignore last two term to approximate $\Delta z$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .