Confusing chain differentiation rule

[frb]

If I have a function f from RxR to R, and a function g from RxR to RxR. What are the partial derivatives of the composition f(g)? I end up multiplying the derivative of f with g, but g is a vector? The partial derivative should have its image in R.

 

[arildno]

Well, the total derivative of g is a matrix, whereas the total derivative of f is a vector. Together, they yield a vector.

Let's take a concrete example:
h(x,y)=f(u(x,y),v(x,y)), g(x,y)=(u(x,y),v(x,y))
We have therefore, for example:
\frac{\partial{h}}{\partial{x}}=\frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{f}}{\partial{v}}\frac{\partial{v}}{\partial{x}}
This is then one of the two partial derivatives of h, the other being differentiation with respect to y.

 

[frb]

Well, the total derivative of g is a matrix, whereas the total derivative of f is a vector. Together, they yield a vector.

Let's take a concrete example:
h(x,y)=f(u(x,y),v(x,y)), g(x,y)=(u(x,y),v(x,y))
We have therefore, for example:
\frac{\partial{h}}{\partial{x}}=\frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{f}}{\partial{v}}\frac{\partial{v}}{\partial{x}}
This is then one of the two partial derivatives of h, the other being differentiation with respect to y.

what does
\frac{\partial{f}}{\partial{u}}
mean? Is the following formula correct?
\frac{\partial{h}}{\partial{x}}(a,b)=\frac{\partial{f}}{\partial{x}}(u(a,b),v(a,b))\frac{\partial{u}}{\partial{x}}(a,b)+\frac{\partial{f}}{\partial{y}}(u(a,b),v(a,b))\frac{\partial{v}}{\partial{x}}(a,b)

 

[HallsofIvy]

what does
\frac{\partial{f}}{\partial{u}}
mean?

It means the partial derivative of f with respect to u, of course. What else could it mean?

Is the following formula correct?
\frac{\partial{h}}{\partial{x}}(a,b)=\frac{\partial{f}}{\partial{x}}(u(a,b),v(a,b))\frac{\partial{u}}{\partial{x}}(a,b)+\frac{\partial{f}}{\partial{y}}(u(a,b),v(a,b))\frac{\partial{v}}{\partial{x}}(a,b)

It doesn't make sense. If f(u(a,b),v(a,b)) makes any sense then f is a function of u and v, not x and y. You must mean \frac{\partial f}{\partial u} not \frac{\partial f}{\partial x}.