How do you find #(dy)/(dx)# given #4x^2+3xy^2-6x^2y=y^3#?

Question

Isabella Ackerman · Accepted Answer

#dy/dx=(-8x-3y^2+12xy)/(6xy-6x^2-3y^2)#

Given - #4x^2+3xy^2-6x^2y=y^3# #8x+[(3xdy/dx2y)+(y^2. 3)]-[(6x^2 .dy/dx)+(y. 12x)]=dy/dx3y^2# #8x+[6xydy/dx+3y^2]-[6x^2dy/dx+12xy]=dy/dx3y^2# #8x+6xydy/dx+3y^2-6x^2dy/dx-12xy=dy/dx3y^2# #8x+6xydy/dx+3y^2-6x^2dy/dx-12xy-dy/dx3y^2=0# #6xydy/dx-6x^2dy/dx-dy/dx3y^2=-8x-3y^2+12xy# #dy/dx(6xy-6x^2-3y^2)=-8x-3y^2+12xy# #dy/dx=(-8x-3y^2+12xy)/(6xy-6x^2-3y^2)#

Ezra Adams · Answer

undefined To find $(dy)/(dx)$ given $4x^2+3xy^2-6x^2y=y^3$, differentiate both sides of the equation with respect to $x$ using implicit differentiation. After differentiating, solve for $(dy)/(dx)$.

$$
\frac{d}{dx} (4x^2+3xy^2-6x^2y)=\frac{d}{dx} (y^3)
$$

$$
8x + 3y^2 \frac{dy}{dx} + 3xy^2 - 6(2xy \frac{dy}{dx} + x^2)=3y^2 \frac{dy}{dx}
$$

$$
8x + 3y^2 \frac{dy}{dx} + 3xy^2 - 12xy \frac{dy}{dx} - 6x^2 = 3y^2 \frac{dy}{dx}
$$

$$
8x - 6x^2 + 3xy^2 = 9y^2 \frac{dy}{dx} - 12xy \frac{dy}{dx}
$$

$$
8x - 6x^2 + 3xy^2 = (9y^2 - 12xy) \frac{dy}{dx}
$$

$$
(8x - 6x^2 + 3xy^2)/(9y^2 - 12xy) = \frac{dy}{dx}
$$