my first derivative with calculus
0includes book list
correction update aug 9 , 2016
( 2x + h ) is Exact for a macro triangle .
( 2x + h ) h becomes an added error , at the infinitesimal level
( 2x ) is Exact at the infinitesimal level
still using equaiton f( x ) =to x^2
let x = 0 , for a point( 0 , 0^2 ) on the blue curve .
(2*0 + h ) =to ( 0 + h )
we know , at the trough of the x^2 curve , the slope really is 0 "flat"
and the h has adds an infintesimal error to the result .
We need to remove h from the slope equation , NOT because h has become insignificant to 2x , but because h is adding an error value to the result .
With circuit analysis , many times , an insignificant value is removed from the equation , leaving us with a usefull but approximate equation .
With the infintesimal derivative , when we remove h , we actually get an Exact equation , and not an approx one .
i mixed conventions ,...
i was thinking of circuit analysis technique when i said that
"h becomes insignificant to 2x , as h goes toward 0"
actually , when x =to 0 , ( 2x + h ) =to ( 0 + h )
we see that h is actually larger then 2x , because 2x is now 0
,...sooo ,... ops ,...
write\develop\build the symbolic generic slope equation
which is the instantaneous rate of a curve .
part 2 , update Aug 10 , 2016
as h goes toward 0 , equation ( 2x + h ) converges to 2x .
( partially thanks to irrational numbers )
( some book's use h as a "delta x" )
All the while , dy and dx still exist as infinitesimals .
Just because ( 2x + h ) converges to 2x , doesn't say that
we let h become 0 . h is still NON-zero .
If h were actually allowed to become 0 , then dy and dx
differentials ( f( a ) - f( b ) ) and ( a - b ) would both become
0 too , and we would have nothing to work with .
Thus we can write : dy/dx =to 2x ,...
where , the ratio dy/dx has a tangable result , namely , 2x ,
while each dy and dx have unspecified infinitesimal values