Question

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?

Answer 1

is this under implicit differentiation?

Answer 2

yes

Answer 3

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

Answer 4

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

Answer 5

why does this have to be implicit differentiation?

Answer 6

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

Answer 7

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

Answer 8

you can also use the chain rule

Answer 9

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

Answer 10

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

Answer 11

http://calc101.com/webMathematica/derivatives.jsp#topdoit

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

Ok I see thank you very much for the help.

Answer 21

welcome