# Quantifying “weight” or “control” of a variable to the value of a function?

Say I have an equation

$$x = f(x)$$

I know that here, the independent variable $x$ controls 100% of the variability of the value of the function. It is the only "knob" that I need to turn to manipulate the value of the function. That is, compared to another function:

$$xy = f(x,y)$$

In this function, $x$ does not control the entire equation. The value of $y$ also controls the equation. I have two "knobs" to control, and each has its independent effect on the function.

My question is, how do I measure how much effect a variable has over a function?

To clarify: say my equation was

$$x + \frac{y}{1000} = f(x,y)$$

Then increasing the value of $x$ by 1 will increase the value of the function by 1. But to get to the same effect using $y$, I have to increase its value by 1000. It seems easy enough to say that $x$ has a weight 1000 times that of $y$.

Similarly, in the first equation $x = f(x)$, $x$ has 100% of the control. In the second equation, each variable has 50% of the control.

But these values I'm saying - about how much "control" a variable has on the function's value - are only intuitive. I cannot prove their truth.

How do I know how much control a variable has over the value of the function, especially in the general case when an equation becomes complex, like when the variable cannot be factored out:

$$x + \frac{y}{x} = f(x,y)$$

Or when the value of the variables are "polluted" by constants or itself:

$$\frac{\sqrt{x + 1000}}{99} + \frac{y}{y+1} = f(x,y)$$

Or when there are more than two variables:

$$x + yz = f(x,y,z)$$

Sounds like the weight is just the partial derivatives with respect to the variable you wish to measure.

$$x+y1000=f(x,y)$$

The derivative with respect to $y$ is $1000=dxf(x,y)/dy$ using the power rule.

The derivative with respect to $x$ is $1=dyf(x,y)/dx$

If the equation were:

$$x^2+1000y^3$$

then respectively

$$3000y^2=dxf(x,y)/dy$$

and

$$2x = dyf(x,y)/dx$$

and the "weight" depends very much on the point at which $x$ is being evaluated, as it can change.

for $$x+y/x=f(x,y)$$, I am hazy on my rules for differentiation, but I looked it up:

for dx/dy we can rewrite:

$$x+y*(x^-1)$$

Y is a constant here. $$dx/dy = 1 + -1*(x^-2)*y$$

Always look how you can rewrite an equation, and also try wolframalpha.com if you dont already know about it.

• Can you explain your derivative notation of $df(x,y)dx$? I've never seen the differential of the dependent and independent variable multiplied with one another rather than divided. And also, these should be partial derivatives, not total derivatives. – Ataraxia Sep 8 '13 at 17:30
• No, thanks, it needs to be edited to include a "/" for division. Right and I also need to fix the notation to be: dy/dx and dx/dy. I've been out of calculus for a couple of years now, sorry. – Mr. AM Sep 8 '13 at 17:33
• Thanks. I knew it had something to do with partial derivatives, but this doesn't fully answer my question. Does this mean that the "weight" I'm looking for is, say I'm looking for the weight of $x$ in $x^2 + 1000y^2$, $\frac{partial derivative of x}{(partial derivative of x) + (partial derivative of y)}$ = $\frac{2x}{2x + 3000y^2}$? – markovchain Sep 8 '13 at 17:38
• No, that's still not correct. If you want to do Leibniz notation, it should be $\frac{\partial}{\partial{x}}f(x,y)$ and $\frac{\partial}{\partial{y}}f(x,y)$. Note that $\frac{d}{dx}f(x,y)$ and $\frac{d}{dy}f(x,y)$ are total derivatives, not partial ones. – Ataraxia Sep 8 '13 at 17:38
• You can also use function notation, which I tend to prefer: $f_x(x,y)$ and $f_y(x,y)$. – Ataraxia Sep 8 '13 at 17:41