Proof of derivative of determinant

1. Nov 5, 2009

unfunf22

1. The problem statement, all variables and given/known data
Given n2 functions fij, each differentiable on an interval (a,b), define F(x) = det[fij] for each x in (a,b). Prove that the derivative F'(x) is the sum of the n determinants, F'(x) = $$\sum_{i=0}^n det(Ai(x)).$$ where Ai(x) is the matrix obtained by differentiating the functions in the ith row of [fij(x)].

2. Relevant equations
To be honest I'm not completely sure what equations would be useful in this proof. I cannot get a good intuition on it.

I suppose Leibniz formula could be handy: http://en.wikipedia.org/wiki/Leibniz_formula_for_determinants

Expansion by minors might be useful as well: http://en.wikipedia.org/wiki/Laplace_expansion

However, that is as far as my intuition will take me.
3. The attempt at a solution
This has been more of a sit and think process than a pen to paper one. I understand what I am supposed to do, but I have not the slightest idea what my intuition to this proof should be. I can't see how the determinant of the matrix would look (since I cannot diagonalize it or anything to make the determinant easy to calculate). I'm not very strong in my understanding of the Leibniz formula for determinants. I understand the idea behind minor's a bit better (delete i-row,j-column and take determinant of what is left). If I could get a feel for what the determinant of [fij(x)] would look like I might be able to see why the derivative would look like F'(x) = $$\sum_{i=0}^n det(Ai(x)).$$

I just need some intuition here, basically, and perhaps a good explanation of the weapons in my arsenal that would help me do this proof.

-Ian

2. Nov 5, 2009

Office_Shredder

Staff Emeritus
Think about writing out the determinant as summing over every possible term $$\Pi f_{ij}$$ where no pair of fij is from the same row or column. Then think about the product rule

3. Nov 5, 2009

unfunf22

So you are saying use leibniz and match its behavior to that of the product rule? Is there not a way to do this inductively?

4. Nov 5, 2009

Office_Shredder

Staff Emeritus
I guess you can. If you expand by minors along a row, take the derivative of that, and then use your inductive step to find the derivative of the minors

5. Nov 5, 2009

unfunf22

Alright. Well I am using Leibniz to informally argue it at the moment. I'll see if that pans out.

6. Nov 5, 2009

unfunf22

I was able to prove it thanks to the general formula for the product rule of n functions. Thanks!