d(x^2)/dx = x

d(x^2)/dx = x

PROOF:

d(x^2)/dx = d(x + x + x + ....... (x times))/dx

= d[x]/dx + d[x]/dx + d[x]/dx....... (x times)

= 1 + 1 + 1 + ....... (x times)

= x

QED

😉

Now for integer (non-positive), rational (non-integer) or real (non-rational) x

Originally posted by TheMaster37
Now for integer (non-positive), rational (non-integer) or real (non-rational) x

😉

I could point out that d/dx is defined with limits, the quotient of the horizontal and vertical displaccement in a graph, with the horizontal displacement going to 0. Since that limit process is kinda abrubt on the integers; 3,2,1,0,0,0,0... you divide by 0 and can get any answer you like.

Assuming the limit process is correctly defined for natural numbers...

How about this: let f(x) = x*[x], where [x] is the integer of smallest modulus which has greater modulus than x.

If x isn't an integer, f'(x) = [x]
OTOH if g(x) = x^2, g'(x) = 2x

So on the set of positive reals which aren't integers, f and g are differentiable and g' - f' tends to infinity as x tends to infinity, but f > g for all x. (*)

This is a demonstration of why your domain must be connected if you want differentiable (and indeed continuous) functions to behave themselves. For example, it's not possible to find two functions which satisfy (*) on all the positive reals.

I'm a litle bit in the dark on what you mean by modulus...[x] is usually the entier-function; [x] is the greatest integer no exceeding x (a sort of rounding down).

Originally posted by THUDandBLUNDER
d(x^2)/dx = x

PROOF:

d(x^2)/dx = d(x + x + x + ....... (x times))/dx

= d[x]/dx + d[x]/dx + d[x]/dx....... (x times)

= 1 + 1 + 1 + ....... (x times)

= x

QED

😉

Originally posted by crec2k
Brillient

Originally posted by crec2k
That was a purposeful mistake.
FYI, myles is fat.