I have been racking my brain on how i should go about resolving:

$\frac{d( n (\frac{g}{n} - \lfloor\frac{g}{n}\rfloor))}{d \lfloor\frac{g}{n}\rfloor}$ For this case replace ${\lfloor\frac{g}{n}\rfloor}$ as ${f(\frac{g}{n})}$ (assume it is continuous)

How can I go about this, when this whole thing seems intertwined. I'm not looking for a partial; but even if I was, I wouldn't know where to start.

{ The original differential was $\frac{d mod(g,n)}{d g//n}$ where

mod(g,n) ==modulus of g wrt n

g//n == floor(g/n)}

  • 2
    $\begingroup$ I can't think of an interpretation of the notation where taking derivatives makes sense. None of these functions are continuous, and the functions in the denominator are piecewise constant. $\endgroup$ – Greg Martin Mar 30 at 7:28

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