Proof of the total differential of f(x,y)?

[Curl]

If I have a smooth, continuous function of 2 variables, z=f(x,y)

I want to show what Δz ≈ (∂z/∂x)Δx + (∂z/∂y)Δy

Most places I've seen call this a definition, but it's not really that obvious. I know that it makes perfect sense geometrically, but I want a little more.

One way I thought of approaching it is to put a tangent plane at the point x0 y0 and show that going along x then along y is like cutting diagonally across to x,y.
Basically I need to show that f(x+Δx ,y+Δy) = f(x+Δx, y) +f(x, y+Δy) - f(x,y).

Unfortunately I'm not good at math, not good at proofs, tired, and a bit busy/lazy :), so I'm calling in the troops. Thanks!

 

[tiny-tim]

Hi Curl!

Try starting with f(x+Δx ,y+Δy) - f(x,y)

= f(x+Δx, y+∆y) - f(x+Δx, y) + f(x+Δx, y) - f(x, y).

 

[HallsofIvy]

Another way of looking at it is this: suppose x and y were functions of some parameter, t.

Then f(t)= f(x(t),y(t)) and, by the chain rule,
\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}
In terms of the differential, we can write that as
df= \frac{df}{dt}dt= \left(\frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}\right)dt= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy
which is now independent of t.

 

[Curl]

Hi Curl!

Try starting with f(x+Δx ,y+Δy) - f(x,y)

= f(x+Δx, y+∆y) - f(x+Δx, y) + f(x+Δx, y) - f(x, y).

hehehe, clever! thanks.
And yes, I've thought of using the chain rule, but at this point we can't prove the chain rule without proving this. So it's like the chicken and the egg.

 

[Hurkyl]

If I have a smooth, continuous function of 2 variables, z=f(x,y)

I want to show what Δz ≈ (∂z/∂x)Δx + (∂z/∂y)Δy

Most places I've seen call this a definition, but it's not really that obvious. I know that it makes perfect sense geometrically, but I want a little more.

Just FYI, it sounds like what you are really asking for is a demonstration differential approximations are good approximations.