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Deriviative by infitessimal multiplication

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Derivative by infinitesimal multiplication

Newton's derivative by infinitesimal addition

The limit when a --> 0 is used to check the slope of a point on a graph.
Calculus uses the addition to derive the slope at a point.

a line f(x)=x, with two points y1, y2

x=y1
x+a=y2

for a number n
if
f(x+n) = f(x)+f(n)

then derivative is

f(x+a)-f(x)    f(x)+f(a)-f(x)        f(a)
-----------  = ------------------  = -----------
(x+a) - x            a                       a

which when evaluated creates indeterminate answer at x=0

lim        f(0)     infnity          0
a-->0 ------- =            or  ------
          0                            0

We could also check a point on a graph with multiplication of tiny a.

if
f(a*x) = a*f(x)

lim        f(a*x) - f(x)     f(x)*(a - 1)        f(x)
a-->0 --------------- =   ------------   =   ------
                a*x - x            x*(a-1)          x

So any function where a constant can be pulled out will have the above
form if multiplication happens.

Also, calling multiplicative derivative "merivative"
because first come first served. :3
(Actually, someone probably thought of this at some point already.)

Right, because for merivitave, f(x)=x (a line) is gonna be wonky.

f(x)=x

Mx(f(x)) = ?

lim    ax - x    lim        x(a-1)    lim        1
a-->0 ------- = a-->0  --------- = a-->0 ------
            ax-x                x(a-1)               1

That limit makes no sense, so we can't have it go like that, so, for f(x)=x
we gotta have an exception. Oh hey, isn't that how Newton's derivitive works?
Gotta have both! So, for a equation with x*f(x), how's it going to work?
Keep it stupid, simple! Better to have an error at the start then the end, eh?

Mx[f(x)]= (x+a)*f(a*x)

1) Mx[f(x)]=x^2

lim        ax^2 - x^2        x^2(a-1)        x^2    
a-->0 ----------------  =  ------------    = ---- = x
            ax-x                    x(a-1)           x


2) Mx[f(x)]=x^n

lim    ax^n - x^n    x^n(a-1)    x^n  
a-->0 ---------- = ------------ = ------- = x^(n-1)
            ax-x            x(a-1)         x      

Similar, but different than calculus where Dx(x^n)= n*x^(n-1)

f(x)=cos(a*x)

using half-angle formula

lim    cos(a*x)-cos(x)    (0.5a*cos(x)^2 - 1) - cos(x)
a-->0 --------------- = ----------------------------
            ax-x                    x(a-1)

lim    cos(x)(0.5a*cos(x)-1) - 1
a-->0 -------------------------
                x(a-1)


lim    cos(x)(0.5a*cos(x)-1)          lim     -1
a-->0 ---------------------     +     a-->0  -----
                x(a-1)                                x(a-1)

     cos(x)(0 - 1)      -1
-----------------   + --------
     x(0-1)             x(0-1)

         cos(x)    1
        ------ + ---
         x          x


f(x)=sin(a*x)

lim    sin(a*x)-sin(x)      lim     0.5*a*sin(x)cos(x) - sin(x)
a-->0 ---------------     = a-->0 --------------------------
            a*x-x                                 x(a-1)


lim     sin(x)(0.5*a*cos(x)-1)    sin(x)(0-1)     sin(x)
a-->0  ----------------------  = ---------------- =    ------
                x(a-1)                       x(0-1)         x



??
y2-y2   sin(x)
----- = ------
x2-x1     x
??

if sin(x)=1
then is related to merivative of cos(x) ?

                    cos(x)  (1)
Mx(cos(x)) = -----  + ----
                     x         x

if cos(x)=0
then it too could be related to merivative of sin(x)

                 (0)     sin(x)
Mx(sin(x) = ---- + -----
                  x        x


f(x)=tan(x)

tan(a*x)-tan(x)  -sec(x)*sin(x-a*x)*sec(a*x)     0
-------------- = ---------------------------------   = ---
a*x-x                     a*x-x                              x

Here tan(x)=0
is due to simplification?
is due to f1(x)-f2(x)=0?
needs fiddling!

Since x+x+x+... = n*x
A multiplicative derivative can either be random, or some constant number of additions.

a = dx+dy+dz+dw+...
b = dx+dx+dx+dx+... = n*dx

x+a = x+a = x+dx+dy+dz+dw+... !~ x+dx
x+b = x+n*dx ~ x+dx

a*x = x*(dx+dy+dz+dw+...) !~ x+dx
b*x = n*dx*x ~ lim dx->0 x+dx

I should check when lim 0 and inf, as they are both indeterminate when additive derivation.


f(x) = x*cos(x)

             (x+a)*cos(a*x) - x*cos(x)
Mx(f(x)) = -------------------------
                             a

            x*cos(a*x)+a*cos(a*x) - x*cos(x)
            = --------------------------------
                             a

                     cos(x)    1
Mx(cos(a*x) = -----  + ---
                     x        x

                x*(cos(x)/x  + 1/x) +a*(cos(x)/x  + 1/x) - x*cos(x)
             = ----------------------------------------------------
                                     a

              cos(x) +1 +a*(cos(x)/x  + 1/x) - x*cos(x)
          = -------------------------------------------------
                                    a

             cos(x)  +1  +a*(cos(x)/x  + 1/x) - x*cos(x)
          = ------    ----  -------------     -------
                a       a      a                   a

              cos(x)   +1 . +(cos(x)/x  + 1/x) - x*cos(x)
          =  ------    ----                    ------
                 a         a                        a

Here, the question of "how small or big is 'a'" is interesting!
If 0, 3 indeterminate terms.
If large, 3 zeroes.

Kinda interesting, since there are 3 ways x*cos(x)=0.
1 linear x=0
2 trig circle cos(x) for x={-1,1}

0*cos(n)=0
1*cos(1)=0
-1*cos(-1)=0

Mx(x*sin(x)) = (x*a)*sin(a*x) - x*sin(x)

            x*sin(a*x) + a*sin(x) - x*sin(x)
         = --------------------------------
                                a

             x*sin(x)/x + a*sin(x) - x*sin(x)
         = --------------------------------
                                a

             x*sin(x)*(1/x -1) + a*sin(x)
         = -------------------     --------
                     a                     a

             x*sin(x)*(1/x -1) + sin(x)
         = -------------------- 
                        a

if

lim a
--> 0  

= f(x)/0 + sin(x)

yet if

lim a
--> infinity  

= f(x)/infinity + sin(x)

= sin(x)
I try to math or something. Me having flunked Calc 3 should be apparent from reading this. Eh. Live and learn!

Needs more thinking!
Comments1
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Zig-zag-zug's avatar
Darn it, DA edited the spacing. :(