infinitessimal limited to calculable functions

The cardioid equation is
f(x,y) = (x^2 + y^2 - ax)^2 = a*a(x^2 + y^2)
with c = (x^2 + y^2)^0.5, f(x,y) = g(x,c) = (c^2 - ax)^2 = a^2*c^2
deriving it by variable t
f'(x,y) = d(c^2 - ax)^2/dt = a^2*c^2
f'(x,y) = 2(c^2 - ax)(2c*dc/dt - x*da/dt - a*dx/dt) = 2a^2*c*dc/dt + 2a*c^2*da/dt
f'(x,y) = (c^2 - ax)(2c*dc/dt - x*da/dt - a*dx/dt) = a*c(a*dc/dt + c*da/dt)
f'(x,y) = (2c*dc/dt - x*da/dt - a*dx/dt) = a*c(a*dc/dt + c*da/dt)/(c^2 - ax)
diff. eq.: 1/dt*[2dc/a - x/a/c*da - dx/c = (a*dc + c*da)/(c^2 - ax)]
f'(x,y)dt = 2/a - x/a/c*da/dc - dx/dc/c = (a + c*da/dc)/(c^2 - ax)
deriving with dx/dc = (dc/dx)^-1 and c = (x^2 + y^2)^0.5
f'(x,y)dt = 2/a - x/a/(x^2 + y^2)^0.5(d(x^2 + y^2)^0.5/da)^-1 - 1/(x^2 + y^2)^0.5(d(x^2 + y^2)^0.5/dx)^-1 = (a + (x^2 + y^2)^0.5(d(x^2 + y^2)^0.5/da)^-1)/(x^2 + y^2 - ax)
f'(x,y)dt = 2/a - x(x + y)/a/(x^2 + y^2) - (x + y)/(x^2 + y^2) = [a + (x + y)/(x^2 + y^2)]/(x^2 + y^2 - ax)
f'(x,y)dt = 2/a - x(x + y)/a/(x - y)/(x + y) - (x + y)/(x + y)/(x - y) = [a + (x + y)/(x + y)(x - y)]/(x^2 + y^2 - ax)
f'(x,y)dt = 2/a - x/a/(x - y) - 1/(x - y) = [a + 1/(x - y)]/(x^2 + y^2 - ax)
f'(x,y)dt = -a[a + 1/(x - y)]/(x^2 + y^2 - ax) - a/(x - y) + 2 = x/(x + y)

This eq. f'(x,y)dt  cannot be integrated.
f'(x,y) is a number producing equation
yet is multiplied by an infinitessimal
a number so small it is near zero
It can be derived or integrated again if the following is done:
since u*v = u + c there is a constant of derivation that here is called n
such that u*dt = u + n

Let's consider f'(x,y)dt.

f'(x,y)*dt = f'(x,y) + n

when lim dt --> 0 then f'(x,y) + n must tend towards 0

so f'(x,y) + n ≈ 0

For algebraic function, n --> infinity as

x^2 + n ≈ 0
(5)^2 + (-24.999...) ≈ 0
(25)^2 + (-625.0001) ≈ 0
and so on

For some functions, this is not the case. One example is cos(x) = f(x), which oscillates between -1 and 1.

cos(x) + n ≈ 0

and so, cos(x)*dt = cos(x) + n ≈ 0

here, n is necessarily between -1, 1 for all real numbers

f(x) = g(u)
sin(x) = u^2 + u
f'(x) = g'(u)
cos(x)*dx/dt = 2u*du/dt + 1*du/dt
cos(x)*(1) = (2u + 1)du/dt
diff. eq. f'(x)dt = cos(x)*dt = (2u + 1)du
cos(x)*dt = cos(x) + n ≈ 0
cos(x) + n = 2u + 1 + m
n = ± 1
cos(x) ± 1 = 2u + 1 + m
cos(x) - 2u - 1 ± 1 = m
let's put u = tan(x)
cos(x) - 2tan(x) - 1 ± 1  = m
checking the bounds with cos(x) = ±1
-2tan(x) - 1 ± 2 = m

m is between ±infinity ± 2

Okay, tan(x) was a toughie.
Let's make it something not oscillating so bigly.

cos(x) - 2u - 1 ± 1 = m
with u = sin(x)cos(x)
cos(x) - 2sin(x)cos(x) - 1 ± 1 = m
at the utmost
± 1 - 2(±0.5) - 1 ± 1 = m
± 3 - 1 = m
The thing (m) bounces between [-4,2]

Trying with u = sin(x)
cos(x) - 2sin(x) - 1 ± 1 = m
± 1 ± 2 - 1 ± 1 = m
± 4 - 1 = m

m is between [-5,3]