Question

How do you find `dy/dx` by implicit differentiation given `x^2+3xy+y^2=0`?

Answer 1

Given: `x^2+3xy+y^2=0`

Differentiate each term with respect to x:

`(d(x^2))/dx + (3d(xy))/dx+ (d(y^2))/dx=(d(0))/dx`

Use the power rule, `dy/dx = nx^(x-1)`, on the first term:

`2x + (3d(xy))/dx+ (d(y^2))/dx=(d(0))/dx`

Use the product rule, `(d(xy))/dx= dx/dxy+xdy/dx = y + xdy/dx` on the second term:

`2x + 3(y + xdy/dx)+ (d(y^2))/dx=(d(0))/dx`

Use the chain rule, `(d(y^2))/dx=2ydy/dx`, on the third term:

`2x + 3(y + xdy/dx)+ 2ydy/dx=(d(0))/dx`

The derivative of a constant is 0:

`2x + 3(y + xdy/dx)+ 2ydy/dx=0`

Distribute the 3:

`2x + 3y + 3xdy/dx+ 2ydy/dx=0`

Move all of the terms that do not contain `dy/dx` to the right:

`3xdy/dx+ 2ydy/dx=-(2x+3y)`

Factor out `dy/dx`:

`(3x+2y)dy/dx=-(2x+3y)`

Divide by `3x+2y`:

`dy/dx=-(2x+3y)/(3x+2y)`

Answer 2

`dy/dx=-(2x+3y)/(3x+2y)`

Explanation:

`"differentiate "color(blue)"implicitly with respect to x"`

`"the term " 3xy" is differentiated using the "color(blue)"product rule"`

#rArr2x+3(x.dy/dx+y.1)+2y.dy/dx=0#

`rArr2x+3xdy/dx+3y+2ydy/dx=0`

`rArrdy/dx(3x+2y)=-2x-3y`

`rArrdy/dx=-(2x+3y)/(3x+2y)`

Answer 3

`(dy)/(dx)=-1/2(sqrt 5 pm 3)`

Explanation:

From `x^2 + 3 x y + y^2=0 -> y = -1/2(sqrt 5 pm 3)x`

then

`(dy)/(dx)=-1/2(sqrt 5 pm 3)`