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dy/dx of y=(sqrt(x)-1)/(sqrt(x)+1)
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\[y=(\sqrt{x-1})/(\sqrt{x+1)}\]
is it?
@mathhelpkl
it's \[y=\sqrt{x}-1/\sqrt{x}+1\]
ok..ok
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Do you knw the answer?
not quite...I'm really close though I think
where are you now on your solution?
I've gotten to \[y \prime=1/(\sqrt{x}+1) \] but I don't think that's quite right
* the bottom is squared
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i am getting \[y'=1/(\sqrt{x}*(\sqrt{x+1})^2)\]
how did you get \[\sqrt{x}\] on the bottom
|dw:1379304969616:dw|
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