Prove Derivative is sum of determinants Given $n^2$ functions $f_{ij}$, each differentiable on an interval (a,b), define $F(x) = det[f_{ij}(x)]$ for each $x$ in $(a,b)$. Prove that the derivative $F'(x)$ is the sum of the determinants, $$ F'(x) = \sum_{i=1}^n det A_i(x),$$
where $A_i(x)$ is the matrix obtained by differentiating the functions in the $ith$ row of $[f_{ij}(x)]$
Yeah, I have no clue what I'm supposed to do.
 A: $$\begin{align}
{F'(x)} & = \frac{d\, \det\,[f_{i,j}(x)] }{d\,x} & \text{ apply Leibniz formula for determinant }\\
& = \dfrac{d\, \sum_{\sigma \in S^n} ({\rm sgn~}\sigma) \prod_{k=1}^{n} f_{k,\sigma _k}(x) }{d\,x}\\
& = \sum_{\sigma \in S^n} ({\rm sgn~}\sigma)\frac{d\, \prod_{k=1}^{n} f_{k,\sigma _k}(x) }{d\,x} & \text { apply derivative of product rule } \\
& = \sum_{\sigma \in S^n} ({\rm sgn~}\sigma)\sum_{i=1}^{n}\left(\prod_{k=1}^{n} \begin{cases}
f_{k,\sigma _k}(x) \text{ if } i \ne k\\
f'_{k,\sigma _k}(x) \text{ if } i = k\\
\end{cases}\right)\\
& = \sum_{i=1}^{n}\sum_{\sigma \in S^n} ({\rm sgn~}\sigma)\left(\prod_{k=1}^{n} \begin{cases}
f_{k,\sigma _k}(x) \text{ if } i \ne k\\
f'_{k,\sigma _k}(x) \text{ if } i = k\\
\end{cases}\right)\\
& = \sum_{i=1}^{n}\det\left[\begin{cases}
f_{k,j}(x) \text{ if } i \ne k\\
f'_{k,j}(x) \text{ if } i = k\\
\end{cases}\right] \\
& = \sum_{i=1}^{n}\det\, A_i
\end{align}$$
A: From
Differentiating the determinant of the Jacobian of a diffeomorphism (don't understand a proof)
we get
$$ \frac{d}{dt} \det(G(t)) = \det(G(t)) \text{trace}[G(t)^{-1} G'(t)] .$$
where $G(t) = [f_{ij}(t)]$ and $t = x$.
Now $\det(G(t)) G(t)^{-1} = H(t)^T$ where $H(t)$ is the matrix of cofactors of $G(t)$:  http://en.wikipedia.org/wiki/Adjugate_matrix.
So
$$ \frac{d}{dt} \det(G(t)) = \sum_{i=1}^n \sum_{j=1}^n H_{ij}(t) G'_{ij}(t) .$$
But by standard methods for computing the determinant http://en.wikipedia.org/wiki/Determinant#Laplace.27s_formula_and_the_adjugate_matrix, $\sum_{i=1}^n H_{ij}(t) G'_{ij}(t) = \det(A_j)$.
This proof works if $G(t)$ is invertible.  If $G(t)$ is not invertible, I think you can follow the second proof I have given here: Metric on an open subset of $\mathbb{R}^d$ and Christoffel symbol of the second kind
