Question

Confirm my answer?
I was asked to find dy/dx of x^2+y^2+3xy=9. I got -2x/(2y+3). Is this correct?

Answer 1

x'^2+y'^2+3(xy)' = 9'

2x x'+ 2y y'+3(x'y+xy') = 0

Answer 2

since dx/dx = x' = 1

2x + 2y y'+3(y+xy') = 0

2x + 2y y'+3y+3xy' = 0

Answer 3

i like it

Answer 4

(-2x-3y) up top

Answer 5

2x + 2y y'+3y+3xy' = 0

2x + 3y + 2y y'+3xy' = 0

2x + 3y + (2y+3x)y' = 0

Answer 6

\[F(x,y) = x ^{2}+y ^{2}+3xy-9=0\]
\[F _{x}=2x+3y\]
\[F _{y}=2y+3x\]
\[dy/dx=-F _{x}/F _{y}=-(2x+3y)/(3x+2y)\]

Answer 7

Thank you

Answer 8

good luck ;)