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

Question

Emma Aldridge · Accepted Answer

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)#

Christopher Agresta · Answer

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

Luke Alvarez · Answer

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

Axel Alvarez · Answer

undefined To find $ \frac{dy}{dx} $ by implicit differentiation for the equation $ x^2 + 3xy + y^2 = 0 $, follow these steps:

1. Differentiate both sides of the equation with respect to $ x $.
2. Use the chain rule where necessary.
3. Isolate $ \frac{dy}{dx} $ on one side of the equation.

The steps are as follows:

1. Differentiating both sides of the equation:

$$ \frac{d}{dx}(x^2) + \frac{d}{dx}(3xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(0) $$

2. Applying the product rule for $ 3xy $:

$$ 2x + 3\frac{dy}{dx} + 3y + 2y\frac{dy}{dx} = 0 $$

3. Rearranging terms and isolating $ \frac{dy}{dx} $:

$$ (3\frac{dy}{dx} + 2y) + (2x + 3y\frac{dy}{dx}) = 0 $$
$$ 3\frac{dy}{dx} + 2y + 2x + 3y\frac{dy}{dx} = 0 $$
$$ 3\frac{dy}{dx} + 3y\frac{dy}{dx} = -2y - 2x $$
$$ \frac{dy}{dx}(3 + 3y) = -2y - 2x $$
$$ \frac{dy}{dx} = \frac{-2y - 2x}{3 + 3y} $$