# Derivative of Multi Variable Equation for Diablo 3 Damage Equation

It has been years since I took calculus and I am trying to figure out if it is possible to calculate the derivative of a mutli variable equation and what the result would be. This equation is from the video game Diablo 3 that represents your damage based on your critical chance and critical damage. I am trying to determine where are the optimal points (or best bang for buck stat) by using derivatives.

x = critical chance, y = critical damage, z = damage output

Equation: $z = 1 + xy$

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There are two variables in $z$, $x$ and $y$. Taking the partial derivative in respect to $x$ (treating $y$ like a constant), we get:

$$\frac{\partial z}{\partial x}=y$$

similarly, the partial derivative in terms of $y$ is

$$\frac{\partial z}{\partial y}=x$$

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+1 for the partial derivatives, And the extrema will be when these guys vanish, which is boring :'( – Alex Nelson Jul 6 '12 at 16:03
Thank you Argon. I read the wiki article and this makes more sense to me. – icu222much Jul 6 '12 at 16:10
@icu222much No problem, glad to help! – Argon Jul 6 '12 at 21:53

As the other answer says, your partial derivatives are $$\frac{\partial z}{\partial x} = y, \quad \frac{\partial z}{\partial y} = x \, .$$

What does this mean in terms of actual gameplay? It means that, if you currently have stats $x=x_0$, $y=y_0$, you should (approximately) value small changes in $x$ by multiplying them by $y_0$, and value small changes in $y$ by multiplying them by $x_0$.

For example, say that at some point in the game your stats are $x=0.5, y=0.25$ (you have a $50$% chance of scoring a critical, and it does one-quarter again as much damage as an ordinary hit). Then, if you're given the option of increasing $x$ by $0.04$ at the cost of decreasing $y$ by $0.01$, you should value this at $0.04(0.25)-0.01(0.5)=0.01-0.005=0.005$. Since this is positive, that option is beneficial.

Since your initial formula for $z$ was fairly simple, this is a little bit superfluous. You could just plug everything into the original $z$-formula. Your initial damage output would be $$z_0=1+(0.5)(0.25)=1.125 \, .$$ Your new damage output would be $$z_1=1+(0.5+0.04)(0.25-0.01)=1.1296 \, .$$ Again, since $z_1>z_0$, we can see that taking this new option is beneficial.

Note that $z_1-z_0=0.0046$: very close to, but not actually equal to, the $0.005$ we computed by using partial derivatives.

Why would you ever use partial derivatives at all, if you could just use the original function to get an exact answer? In this case, I would say you probably shouldn't unless you're in a big hurry. If $z$ were more complicated, though, it might save you a bunch of time, especially if you've got a bunch of different possible changes to try from the same starting point. Since all you care about is which change leads to the biggest (positive) shift in $z$, you won't need to worry that the your partial-derivative computations are approximations unless you have two different potential options that are really close in value.

One thing to be careful about: since $x$ is a probability, presumably it's capped at $1$ (or maybe at some lower value if D3 doesn't like the idea that all your hits could be criticals). So your formula for $z$ probably isn't universal. But it's pretty obvious what you should do when it breaks down (maximize $y$ at all costs, stop worrying about $x$).

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Thank you Micah for your detailed explanation :) – icu222much Jul 6 '12 at 16:43