Find dy/dx. y=(sqrt{x}-1)/(sqrt{x}+1)

Question

valpey · Answer

http://www.wolframalpha.com/input/?i=derivative+of+y%3D%28x+%5E%7B1%2F2%7D-1%29%2F%28x%5E%7B1%2F2%7D%2B1%29

anonymous · Answer

Can you show me the steps or tell me what I am doing wrong?

valpey · Answer

$$v = x^{1/2} - 1:: dv = (1/2)x^{-1/2}$$

valpey · Answer

Similarly if 
$$u = x^{1/2}+1 :: du = (1/2)x^{-1/2}$$

lalaly · Answer

use the quotient rule
$$y=\frac{\sqrt x-1}{\sqrt x+1}$$
$$f(x)=\sqrt x-1$$$$g(x)=\sqrt x+1$$
$$\frac{dy}{dx}= \frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$$

anonymous · Answer

Yes.. I believe that is what I have? I'm not sure how to simply/condense it.

valpey · Answer

Oops I have my u and v backward. But it looked like you missed the exponent = -1/2 in your derivatives.

anonymous · Answer

Ohh okay! Thanks :) But I am still not sure how to simplify it. |dw:1315987107885:dw| Will it be something like this?