# Confusing chain differentiation rule

1. Aug 9, 2008

### 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.

2. Aug 9, 2008

### 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.

3. Aug 9, 2008

### frb

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

Last edited by a moderator: Aug 9, 2008
4. Aug 9, 2008

### HallsofIvy

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

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 $$\frac{\partial f}{\partial x}$$.