Question

given the implicit relation (x^2)-2xy+(y^3)=0, used partial derivative to calculate dy/dx

Answer 1

\[dy/dx=-F _{x}/F _{y}\]

Answer 2

confuse

Answer 3

\[dy/dx=-(2x-2y)/(-2x+3y ^{2})\]

Answer 4

how did u get get rid of the 0

Answer 5

to find this derivative you have to use implicit diferentiation , or to say it other way, make derivative of x considering y a function of x.\[F _{x}+F _{y}*y'=0\]
where y' =dy/dx
now pass The Fx to left and devide everything by Fy and you got it

Answer 6

can u show it step by step

Answer 7

this is step by step

Answer 8

with (x^2)-2xy+(y^3)=0

Answer 9

\[F _{x}=2x-2y\]
\[F _{y}=-2x+3y ^{2}\]
later just use the formula:
\[dy/dx=-F _{x}/F _{y}=-(2x-2y)/(-2x+3y ^{2})\]

Answer 10

thanks