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.
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.