- #1

- 155

- 7

(a+b)^0.5

I said as far as I know, there is no way, but then the kid said:

f'(x^0.5)= ( (x+h)^0.5-x^0.5 ) / h = 0.5 x ^ -0.5;

(x + h)^0.5 = h(0.5 x ^ -0.5) + x^0.5;

and then, he proposed that x and h can be replaced by other variables, such as a and b, and the reason why the equation (a + b)^0.5 = b(0.5 a ^ -0.5) + a^0.5 is false is because h in the original equation approximated zero, and any term with h in it at the original derivative is removed, therefore, the "actual" way (a+b)^0.5 should be expanded is:

b(0.5 a ^ -0.5) + a^0.5 + c; where c is a series of functions in terms of a and b

I suspect that some of his logic is fallacious when doing the algebra to the derivative, because the equality is only true with h-->0, then again I am not very far into the logic of maths myself, so is there a way to demonstrate that this is wrong? I know the general formula for binomial expansion with integer exponents, but I don't have the proof, so maybe that will help if you can give it to me too.