## anonymous 4 years ago 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?

1. anonymous

is this under implicit differentiation?

2. anonymous

yes

3. anonymous

whoops btw y' =-sqrt(y)/sqrt(x). sorry bout that.

4. anonymous

gah, I dunno if i have the energy for this right now. wolfram might have a step by stepper

5. TuringTest

why does this have to be implicit differentiation?

6. anonymous

These are the instructions I am given. Is there another way to do it?

7. anonymous

it doesn't Have to be, It just seems like a problem they would ask under that section

8. anonymous

you can also use the chain rule

9. anonymous

by manipulating the equation . here i'll get a computer to do it for you :P

10. 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

11. anonymous
12. anonymous

dang it didn't carry the answer. well type (1-x^(1/2))^2 into the function box

13. anonymous

it will give the first and second derivatives with steps but it's not implicit, just chain rule

14. anonymous

it will give the first and second derivatives with steps but it's not implicit, just chain rule

15. anonymous

it will give the first and second derivatives with steps but it's not implicit, just chain rule

16. TuringTest

$y''=-\frac12y^{-1/2}x^{1/2}y'+\frac12y^{1/2}x^{-3/2}$sub in the other guys

17. anonymous

sorry i dunno what's wrong with my computer right now, if it's sending the messages multiple times.

18. anonymous

turing, do you sub y' back into the equation for y''?

19. TuringTest

why not? It'll be messy algebraicly, but after it sub in for y again it'll be true...

20. anonymous

Ok I see thank you very much for the help.

21. TuringTest

welcome