Question

Find dy/dx if x^2+y^2+12xy=0.​

Answer 1

Answer:

a hyperbola

Step-by-step explanation:

dy

=

2xy

x

2

+y

2

+1

2y

dx

dy

=x+

x

y

2

+

x

1

2y

dx

dy

−

x

y

2

=x+

x

1

y

2

=u=2y

dx

dy

=

dx

du

dx

du

−

x

u

=x+

x

1

dx

du

+Pu=Q

x

1

dx

du

−

x

2

u

=1+

x

2

1

∫d(

x

u

)=∫(1+

x

2

1

)dx⇒

x

u

x−x

−1

+C

⇒

x

y

2

=x−

x

1

+C

⇒y

2

=x

2

−1+x

Answer 2

Answer:

$\frac{dy}{dx} = \frac{-2x-12y}{2y+12x}$

Step-by-step explanation:

I think this is an implicit function,,,,here's my explanation I really can't recall what was the precise solving :)

$x^{2}+y^{2}+12xy=0$

so the derivative of $x^{2}$ is 2x,

the derivative of $y^{2}$ is 2y$\frac{dy}{dx}$ with respect to x,

and for the 12xy is I did use product rule,  and the product rule is

uv = udv+vdu read as v times u = u times the derivative of v plus v times the derivative of u, here 12x is u and y is v, so

the derivative of a variable is 1 so y=1$\frac{dy}{dx}$ (if you are working with implicit function just remember to put dy/dx to the derivative of y with respect to x ),

so , 12x(1$\frac{dy}{dx}$) + y(12), because the derivative of 12x is just 12, for this part I forgot the explanation but it's pretty much simple,,,heheh

and here we have, 2x + 2y$\frac{dy}{dx}$ + 12x

2y$\frac{dy}{dx}$ + 12x

then factor out the $\frac{dy}{dx}$,

$\frac{dy}{dx}$[2y + 12x] = - 2x -12y

then to to cancel is division thing (I suppose)

$\frac{dy}{dx}$ [$\frac{2y+12x}{2y+12x}$] = $\frac{-2x-12y}{2y+12x}$,

and finally the $\frac{dy}{dx} = \frac{-2x-12y}{2y+12x}$,,,,,

hopefully this helps, remember I am still a student so if there is/are any

wrong about the answer I provide please and please correct, thank you :)