Question

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

Answer 1

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

Answer 2

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

Answer 3

\[v = x^{1/2} - 1:: dv = (1/2)x^{-1/2}\]

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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