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I want to know if this derivative is correct. I have derived this but not sure if this is correct. I think it is but just to confirm

F= A-(B/C)*D where A,B,C and D are square matrices  dF/dx(partial derivative) = d(A-(B/C)*D)/dx  ----- deriving the final result will be -----------  dA/dx - [(dB/dx - B*inv(C)*dC/dx)/C]*D - (B/C)*dD/dx 
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    Yes, the matrices are functions of$x$and it is B*inv(C)2012-05-14

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In general, if A and B are square matrices that depend on $x$, the multiplication rule applies: $\frac{d(A B)}{dx} = \frac{dA}{dx} B + A \frac{dB}{dx} $

Applying this, you result follows directly (but, please don't write $B/C$ instead of $B C^{-1}$)

$\frac{d(A -B C^{-1} D)}{dx} = \frac{dA}{dx} - \frac{dB}{dx} C^{-1} D - B \frac{d \, C^{-1}}{dx}D - B C^{-1} \frac{dD}{dx} $

What remains is to compute $\frac{d \, C^{-1}}{dx}$. But deriving $C C^{-1} = I $ we get

$ \frac{dC}{dx} C^{-1} + C\frac{d \, C^{-1}}{dx } = 0 \; \Rightarrow \; \frac{d \, C^{-1}}{dx} = - C^{-1}\frac{dC}{dx}C^{^-1} $

which (check) is the generalization of $\frac{d (y^{-1})}{dx} = - \frac{1}{x^2}\frac{dy} {dx}$

Finally

$\frac{d(A -B C^{-1} D)}{dx} = \frac{dA}{dx} - \frac{dB}{dx} C^{-1} D + B C^{-1}\frac{d \, C}{dx} C^{-1}D - B C^{-1} \frac{dD}{dx} $

which, fixing notation, coincides with your result.

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    @rajansthapit: no they aren't (now) :-) (i have missed the initial minus)2012-05-14