# Jacobian when there's a multivariate function inside it

1. Feb 3, 2015

### gummz

1. The problem statement, all variables and given/known data

differentiate the function F(x,y) = f( g(x)k(y) ; g(x)+h(y) )

2. Relevant equations

Standard rules for partial differentiation

3. The attempt at a solution

The Jacobian will have two columns because of the variables x and y. But what then? f is a multivariate function inside the Jacobian!

2. Feb 3, 2015

### BvU

So on top of the standard rules you get the chain rule.
Show some attempt at solution and help is on the way.

To demo my ignorance: Differentiating gives two columns, but one row only, right ?
Is there a significance in the ";" ? You write F ( x , y ) -- a notation which I am also familiar with -- , but then you write f ( u ; v )

3. Feb 3, 2015

### Ray Vickson

Do you mean $F(x,y) = f(u,v)$, where $u = g(x) k(y)$ and $v = g(x) + h(y)$? If so, just apply the chain rule for derivatives. You need to express the answers in terms of the functions $f_1, f_2$, where $f_1(u,v) \equiv \partial f(u,v)/\partial u$ and $f_2(u,v) \equiv \partial f(u,v) / \partial v$.

4. Feb 3, 2015

### Brian T

Consider the partial derivatives that make up the derivative matrix. It should be a 2x2, you have two functions, and take the derivative of both functions wrt x or wrt y.