.
Right... but if the equation is separable...
>>16869623What equation?Differential operators don't necessarily need to be related to differential equations.They are a data type with a fancy multiplication rule.
>>16869618But isn't it derived from the limit formula which is a fraction?
>>16869650should move the infinity limit alongside dx then
>>16869618I just treat it like a fraction anyways. Problem?
Only the d part is the operator
>you can separate variables but this totally isn't cross multiplying the thing that's written as a fraction which isn't a fraction!>What do you mean you want to take the finite difference, cross multiply, and then pass the limit? You can't do that!!!What the actual fuck is their problem
>>16869658This. To emulate the behavior of d/(dx) using the total derivative just requires the restriction dx = 1, dy = 0, dz =0, etc.If you really must use the fraction operator at least write it properly as (dx)\dWriting d/(dx) gets the chain rule wrong when changing variables.d/(dx) becomes d/(dy)*(1/g'(y)) = -g''/g'^2 + (1/g')d/(dy) after substitution x = g(y)(dx)\d becomes (1/g')*((dy)\d) which is the right answer.
>>16869618eh, just use non-standard calculus
>>16869618Well it sure fucking looks like a fraction.
>>16869650Yeah but when I make something dependent on a limit I automatically make dx basically 0. If you act as if you can just rearrange [math]\frac{d}{dx}f(x)=\lim_{h\rightarrow 0}\frac{(f(x+h)-f(x))}{h}=g(x)[/math] into [math]\lim_{h\rightarrow 0}{(f(x+h)-f(x))}=g(x)h[/math] you'd have to explain to me how thats actually useful, since h just goes to zero telling us that f(x)-f(x)=0g(x) which is 0 = 0 which is obvious. Not much can be solved from that.
