Question

help me please.... really don't know....
find dy/dx at point (1,2) of the implicit function (x^2+y^2)^2=25/4xy^2

Answer 1

stuck...
\[=2(x^2+y^2)(2x+2y.\frac{ dy }{ dx })=\frac{ 25 }{ 4 }.y^2+2y.\frac{ dy }{ dx }.\frac{ 25 }{ 4 }x\]

Answer 2

what about this ...?
but im stucking then.. =.=

Answer 3

ok ok i'll try now

Answer 4

You can always just substitute the point and just juggle some numbers to isolate dy/dx

Answer 5

\[\frac{ (x^2+y^2)^2 }{ xy^2 }=\frac{ 25 }{ 4 }\]

Answer 6

then expand ?

Answer 7

and differentiate both side ?

Answer 8

and then take differentiate both sides

Answer 9

huh ?? i dont get it . ok this is the original question.. 2)

Answer 10

really ?? im so sorry

Answer 11

ok so my abobe steps just now is right ?

Answer 12

above*

Answer 13

okkkk wait :)

Answer 14

What are you doing? That second post looks correct

Answer 15

But all you have to do at that point is substitute the point (2,1) for x and y then solve for dy/dx

Answer 16

(1,2) sorry

Answer 17

What whole thing. You are left with a bunch of numbers and dy/dx standing around. Then you isolate it.

Answer 18

I mean on the second post where he mentioned he is stuck. All that is needed is to substitute the point (1,2) for x and y and then solve for dy/dx which appears in the equation

Answer 19

\[2(x^2+y^2)(2x+2y.\frac{ dy }{ dx })-\frac{ 25 }{ 4 }y^2+2y.\frac{ dy }{ dx }.\frac{ 25 }{ 4 }x=0\]

Answer 20

you mean this ?

Answer 21

(x^2+y^2)^2=(25/4)xy^2
2(x^2+y^2)(2x+2yy')=(25/4)y^2+(25/4)2xyy'
x=1, y=2
2(1^2+2^2)(2*1+2*1+2*2*y')=(25/4)2^2+(25/4)2*1*2*y'
Solve for y'

Answer 22

haha ok ok im wrong. sorry im not good in math -.-

Answer 23

What is the difference. I can't tell

Answer 24

What is happening here? I'm confused

Answer 25

thank you to both of u .. all of these really help me ^^ and btw im bot a boy.

Answer 26

not*