Cissoid curve problem

[kaelbu]

I'm not entirely certain if this is the correct message board to post this on, but I figure this class is called the History of the Calculus, so it seems like the best fit.
I actually have the solution to the problem, thus my question pertains to why this solution works, my questions are bold.
The problem: the cissoid is the curve \(\displaystyle y=x^{3/2}(1-x)^{-1/2}\) (0<x<1) Prove that y has one derivative at x= 0 and not two.

The solution: take the derivative of y which gives you \(\displaystyle y'=(1/2)x^{1/2}(1-x)^{-3/2}(3-2x)\) for x>0 (why only for x>o?)
then \(\displaystyle {y(x)-y(0)}/x = y/x = x^{1/2}(1-x)^{-1/2}\) which goes to 0 as x goes to 0. (how is this "taking the derivative"?)
finally \(\displaystyle y'/x = (1/2)x^{-1/2}(3-2x)\) which goes to infinity as x goes to zero so y''(0) does not exist (how is this taking "the second derivative"?)

Thanks!

 

[tkhunny]

The first. It's defined in the first place on 0 < x < 1. Perhaps it is excessive to state x > 0 all overa again. On the other hand, this may emphasize that we've only one side of "at zero" so the derivative should be viewed with suspicion.

The last two questions are simple and the same. Perhaps rewrite the denominator "x - 0" and see if you feel better about it.

 

[kaelbu]

hmm... I see what you are saying. I guess I understand why this way of taking the derivative works, but I don't understand how it is possible that both \(\displaystyle {y(x)-y(0)}/x=y/x=x^{1/2}(1-x)^{-1/2}\) and \(\displaystyle y'=(1/2)x^{1/2}(1-x)^{-3/2}(3-2x)\) can both be the derivative of the original function. Is this because because the second derivative is undefined at zero?
I kind of feel like I'm missing something really obvious here :-/

 

[tkhunny]

No, no. You're not seeing it. What's the definition of the derivative? There are a couple of ways to write it.

Here's a piece of it.

\(\displaystyle \frac{f(x+h) - f(x)}{(x+h)-x}\)

In this formula, if we're around x = 0, what is "h"? It is the displacement from x=0. Backing up to the x-axis, what are the coordinates of any point that isn't the Origin?