# Homework Help: Z = xy, dz/dx = delta z/ delta x, no idea why

1. Dec 3, 2015

### jrm4496

1. The problem statement, all variables and given/known data
Ok this isn't really homework just something I came across and am confused by. I came across a function that looks similar to z = xy and I found that delta_z / delta_x = dz/dx which is really weird to me. This was really strange to me so I checked to see if z = x²y would also be the same and it wasn't which I expected. I don't know how to explain why z = xy has this sort of property where delta_z / delta_x = dz/dx. Sorry if this is a sort of weird question.

2. Relevant equations
For example let's say z = xy, x = 5, y = 2, z = 10. We increase x to 6 then z = 12.
delta_z / delta_x = dz/dx = y => delta_z = y*delta_x = 2*1 = 2. No idea why this is true though it should be like a linear approximation.

3. The attempt at a solution
So I was thinking it's true in the same way that for y=mx+b, delta_y/delta_x = dy/dx, now I don't know what to call these type of functions.

2. Dec 3, 2015

### BvU

Hello jrm,

With this function, you can see that if you keep y constant, then z = constant * x, the equation of a straight line.

3. Dec 3, 2015

### HallsofIvy

First, what do you mean by "similar to z= xy"? Was the function z= xy or not? Second, since you are talking about "delta z/delta x" and "dz/dt", what happens to y? If you are treating y as a constant, then z is a linear function of x so of course you get that property. And you call those functions "linear" functions!

4. Dec 3, 2015

### jrm4496

Ah so this function is linear? I looked online and it showed linear multivariable functions have the form f(x1,x2,x3,...) = a1x1 + a2x2 + a3x3 + ... So I didn't want to call this type of function linear since it does not have this form. By similar I mean it looks like z = (1+y)(A+Bx). Sorry I made a mistake in my first post. It should have been y = 10, and increases to y = 12, not z. I guess a follow up question would be anything of the form f(x1,x2,x3,...) = x1*x2*x3... would have this behavior if you kept all other variables constant when taking a derivative with respect to one variable and it would be called linear?

Thanks for the responses!

5. Dec 3, 2015

### BvU

You found the correct criterion for linearity. That doesn't exclude the possibility that the intersection of the 3D graph of your function with any plane y = constant (or, alternativly, x = constant) is a straight line.
Note that for functions of more than one variable we use partial derivatives: Function $\ z = f(x,y)\$ has partial derivatives $\ \partial z\over \partial x\$ and $\ \partial z\over \partial y\$.

$\ \partial z\over \partial x\$ is a function of x and y obtained by taking the the usual derivative wrt x while keeping y constant. In your $z = xy$ example $\ {\partial z\over \partial x} = y\$.

And in your z = (1+y)(A+Bx) $\ {\partial z\over \partial x} = B(1+y)\$, which still is a straight line -- hence the $\ {\Delta z\over \Delta x} = {\partial z\over \partial x}\$.

--

Picture shows intersections with y = -2 is a straight line. So is intersection with x = 2 -- and any plane with x = constant or y = constant.

6. Dec 3, 2015

### jrm4496

Ah, that clears things up for me. Thanks!