Question

Find the derivative y' of the function defined implicitly by:

Answer 1

\[\sqrt{x+y}+\sqrt{xy}=6\]
now this is what i have so far:
\[(\frac{ 1 }{ 2 }(x+y)^\frac{ -1 }{ 2 }*y'+\frac{ 1 }{ 2\sqrt{xy} }(1*y+y'*x)=0\]

so then i get:
\[\frac{ (x+y)^\frac{ -1 }{ 2 }*y'*\sqrt{xy} }{ 2\sqrt{xy} }+\frac{ y }{ 2\sqrt{xy} }+\frac{ xy' }{ 2\sqrt{xy} }=0\]

Answer 2

not sure if i have the third step done correctly, and not sure where to go from here.

Answer 3

(x+y)' = (1+y')

Answer 4

\((\frac{ 1 }{ 2 }(x+y)^\frac{ -1 }{ 2 }*(1+y')+\frac{ 1 }{ 2\sqrt{xy} }(1*y+y'*x)=0\)

Answer 5

i think ur correct

Answer 6

that's what i got (besides the fact that i missed the one lol)

Answer 7

so then i made my first function over 2squrtxy

Answer 8

just by multiplyinh it by \[\sqrt{xy}\]

Answer 9

i just need to know how to further simplify

Answer 10

no take derivative for that one too

Answer 11

Yeah, the third step is correct. Just find y' now and it's done

Answer 12

and i crossed multiplied so now i have no denominators to just have: \[(x+y)^\frac{ -1 }{ 2 }*(1+y')*\sqrt{xy}+y+y'+xy'=2\sqrt{xy}\]

Answer 13

\(\frac{ (x+y)^\frac{ -1 }{ 2 }*(1+y')*\sqrt{xy} }{ 2\sqrt{xy} }+\frac{ y }{ 2\sqrt{xy} }+\frac{ xy' }{ 2\sqrt{xy} }=0 \\ \large \implies \frac{(x+y)^{-1/2}(1+y')\sqrt{xy}+y+xy'}{2\sqrt{xy}}=0\)

Answer 14

then just do numerator =0 and isolate y'

Answer 15

thats what i had, so then i cross multiplied to get what i just wrote above.

Answer 16

isn't right side = 0 ?

Answer 17

i can't factor out y' though, so do i just divide by everything?

Answer 18

well i cross multiplied to get rid of my denominator.

Answer 19

not sure what to do next!

Answer 20

\(\frac{ (x+y)^\frac{ -1 }{ 2 }*(1+y')*\sqrt{xy} }{ 2\sqrt{xy} }+\frac{ y }{ 2\sqrt{xy} }+\frac{ xy' }{ 2\sqrt{xy} }=0 \\ \large \implies \frac{(x+y)^{-1/2}(1+y')\sqrt{xy}+y+xy'}{2\sqrt{xy}}=0 \\ \huge \implies (x+y)^{-1/2}(1+y')\sqrt{xy}+y+xy'=0 \)

Answer 21

\(\large \implies (x+y)^{-1/2}(1+y')\sqrt{xy}+y+xy'=0\)

Answer 22

why did you get rid of your denominator?

Answer 23

see, right side =0
a/b=0
a=0

Answer 24

that confuses me that i just drop the denominator ... and i still can't isolate y' :/