can someone help me with this math problem?

Question

I have to differentiate implicityly to find dy/dx for the following equation

x^2 y^2 - x^3 y^4 = 3x

what?

Answer 1

x^2 y^2 - x^3 y^4 = 3x

I'm going to do each piece separately.

x^2 y^2
= (2x)(y^2) + (x^2)(2y dy/dx)

x^3 y^4
= (3x^2)(y^4) + (x^3)(4y^3 dy/dx)

3x
= 3

So, now you have:
(2x)(y^2) + (x^2)(2y dy/dx) - [(3x^2)(y^4) + (x^3)(4y^3 dy/dx)] = 3

Simplify and solve for dy/dx

2xy^2 + (2x^2y)(dy/dx) - (3x^2 y^4) - (4x^3y^3)(dy/dx) = 3

(2x^2 y)(dy/dx) - (4x^3 y^3)(dy/dx) = 3 - 2xy^2 + 3x^2 y^4

(dy/dx)(2x^2 y - 4x^3 y^3) = 3 - 2xy^2 + 3x^2 y^4

dy / dx = [3 - 2xy^2 + 3x^2 y^4] / (2x^2 y - 4x^3 y^3)

Answer 2

dy/dx means "take the derivative with respect to x"

so...

x^2 y^2 - x^3 y^4 = 3x

becomes

x^2 y^2 - x^3 y^4 - 3x = 0

which becomes

2x y^2 - 3x^2 y^4 - 3 = 0

Answer 3

2xy^2+x^2*2y*y´-3x^2y^4-x^3*4y^3*y´=3
(2x^2y-4x^3y^3)*y´= 3-2xy^2+3x^2y^4
so
dy/dx =(3-2xy^2+3x^2y^4)/(2x^2y-4x^3y^3)

Answer 4

2xy^2+2x^2yy'-3x^2y^4
-4x^3y^3y'=3
Solving for y'
y'=[3-2xy^2-3x^2y^4]/
[2x^2y-4x^3y^3]

Answer 5

2xy^2+2x^2ydy/dx-3x^2y^4-4x^3y^3dy/dx=3
(2x^2y-4x^3y^3)dy/dx=3-2xy^2+3x^2y^4

dy/dx=(3-2xy^2+3x^2y^4)/(2x62y-4x^3y^3)