Hi,
In my view the fallacy that Anil has pointed is indeed quite legit (That's hacking for you ;-) but differentiation at large is supposed to be applied on what is called a function. Such as
f(x) = x^2 . This function maps a certain input 'x' to certain output. In our case the function is a square function. Which applies to every number (real and imaginary) on a co-ordinate system. This means that for every value of input 'x' there is a output 'x^2'.
It is also convenient to view the left hand side as a variable 'y' and call that variable a dependent variable and the variable 'x' on the right hand side as independent variable.
So differentiation makes sense when a function is defined such as a "square" and operate that function on a independent variable 'x' and study the effect of this operation by changing the input and observing the change in the output.
d/dx(f(x)) = [f(x1 + d) - f(x1)] / [(x1 + d) - x1] ; /*Where 'd' is the incremental change in the input*/
But Anil has attacked the very definition of the function "square" even before it is operated on a dependent variable 'x'.
The statement x^2 = x + x + x + x + x + x ...x times is just the definition of the function square. And 'x' in this context just an arbitrary identifier used to define the function and not the independent variable 'x' yet. But when I take this function, operate it on a independent variable 'x' I am defining a system f(x) such that f(x) = x^2. This means to say that we have a system that follows a certain rule(relation or function) and the rule is f(x) = x^2 and if we'd like to study this system's dynamic response when inputs are subject to change, we differentiate and say that for a function f(x) = x^2 when we change the input from x to x+d, change in the output will be twice the initial value x for very low values of 'd' (limit d tends to 0). It will be (2x+d) if we don't don't take the limits. Lets study this.
x 0 1 2 3 4 5 6 7 8 9
f(x) = x^2 0 1 4 9 16 25 36 49 64 81
d/dx(f(x))= 2x 0 2 4 6 8 10 12 14 16 18
let x1 = 3, d = 3 ==> x1+d = x2 =6
f(x1) = 9
f(x2) = 36
difference in output = f(x2) - f(x1) = 36 - 9 = 27
difference in input d = 3
therefore differentiation will yield 27/3 = 9
which is also equal to 2*3 + 3 (2*x1+d).
We can see that as we make 'd' term as close to zero as possible, the differentiation will yield twice the initial value of input for a small incremental change.
In summary, calculus makes sense when there are relations and not when they are applied on random equations.
Phew,
-Daya.
--
There is no secret ingredient.