Partial differentiation: prove this general result

[unscientific]

Homework Statement

The function f(x,y,z) may be expressed in new coordinates as g(u,v,w). Prove this general result:

The Attempt at a Solution

df = (∂f/∂x)dx + (∂f/∂y)dy + (∂f/∂z)dz

dg = (∂g/∂u)du + (∂g/∂v)dv + (∂g/∂w)dw


df = dg since they are the same thing?

but the components dx, dy, dz, du, dv, dw are different so i can't equate them...

 

[pasmith]

The Attempt at a Solution

df = (∂f/∂x)dx + (∂f/∂y)dy + (∂f/∂z)dz

dg = (∂g/∂u)du + (∂g/∂v)dv + (∂g/∂w)dw


df = dg since they are the same thing?

Yes. Here x, y and z are to be regarded as functions of u, v and w, so you need to work out what dx, dy and dz are in terms of du, dv and dw and substitute those into the equation for df.

 

[unscientific]

Yes. Here x, y and z are to be regarded as functions of u, v and w, so you need to work out what dx, dy and dz are in terms of du, dv and dw and substitute those into the equation for df.

how can we assume that x = x(u,v,w) and y = y(u,v,w) and z = z(u,v,w)?

 

[MrWarlock616]

how can we assume that x = x(u,v,w) and y = y(u,v,w) and z = z(u,v,w)?

Since f(x,y,z) may be expressed in new coordinates as g(u,v,w), it means that the function takes a different form when x, y and z are replaced by u, v and w. For example, the function z(x,y)=x+iy (rectangular) becomes t(r, θ)=r(cosθ+isinθ) (polar) when x is replaced by rcosθ and y is replaced by rsinθ. Thus x and y are both functions of r and θ.