Question

Differentiation problem, please help!

Answer 1

Yes

Answer 2

Suppose \[\sqrt{x}+\sqrt{y}=1\]. Find y''

Answer 3

I found f' already, which I got \[\frac{ -\sqrt{y} }{ \sqrt{x} }\] for, is that right so far?

Answer 4

Yes But Hold On Im Working It

Answer 5

Yes. You are right @stottrupbailey

Answer 6

Thank you :) So I'm confused about the next step though

Answer 7

What do you have to do next?

Answer 8

quotient rule, right?

Answer 9

yes. do you have to find f''?

Answer 10

yes

Answer 11

Did you try?

Answer 12

yes, this is what I have so far\[\frac{ \frac{ -\sqrt{x} }{ 2\sqrt{y} }\frac{ dy }{ dx } +\frac{ \sqrt{y} }{ 2\sqrt{x} } }{ x }\]

Answer 13

I'm just confused about getting dy/dx by itself because this isn't an equation, so I can't subtract and multiply stuff over.

Answer 14

substitute
\[\frac{dy}{dx}=-\frac{\sqrt y}{\sqrt x}\]

Answer 15

Oh, I see, okay just a sec

Answer 16

so is the answer \[\frac{ \sqrt{x}+\sqrt{y} }{ 2x \sqrt{x} }\] ?

Answer 17

I get this finally \[\frac{ d^{2}y }{ dx^{2}}=\frac{ -x^{\frac{ -3 }{ 2 }}-2x^{\frac{ -1 }{ 2 }} }{ y^{\frac{ -3 }{ 2 }} }\]

Answer 18

@stottrupbailey yes you are right

Answer 19

Thanks so much ash2326! :)

Answer 20

you're welcome :D