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

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
Homework Helper
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
Homework Helper
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.

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.

Last edited:
Hurkyl
Staff Emeritus