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

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

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