Find y'' of \[\sqrt{x}+\sqrt{y}=1\]. I found f'= \[-\sqrt{y}/\sqrt{x}\]. I can't figure out where to go from here. Any help?

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Find y'' of \[\sqrt{x}+\sqrt{y}=1\]. I found f'= \[-\sqrt{y}/\sqrt{x}\]. I can't figure out where to go from here. Any help?

Mathematics
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is this under implicit differentiation?
yes
whoops btw y' =-sqrt(y)/sqrt(x). sorry bout that.

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gah, I dunno if i have the energy for this right now. wolfram might have a step by stepper
why does this have to be implicit differentiation?
These are the instructions I am given. Is there another way to do it?
it doesn't Have to be, It just seems like a problem they would ask under that section
you can also use the chain rule
by manipulating the equation . here i'll get a computer to do it for you :P
yes implicit differentiation is subject of the section. ty for the help. I used wolfram but I wasn't sure that i had the correct input
http://calc101.com/webMathematica/derivatives.jsp#topdoit
dang it didn't carry the answer. well type (1-x^(1/2))^2 into the function box
it will give the first and second derivatives with steps but it's not implicit, just chain rule
it will give the first and second derivatives with steps but it's not implicit, just chain rule
it will give the first and second derivatives with steps but it's not implicit, just chain rule
\[y''=-\frac12y^{-1/2}x^{1/2}y'+\frac12y^{1/2}x^{-3/2}\]sub in the other guys
sorry i dunno what's wrong with my computer right now, if it's sending the messages multiple times.
turing, do you sub y' back into the equation for y''?
why not? It'll be messy algebraicly, but after it sub in for y again it'll be true...
Ok I see thank you very much for the help.
welcome

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