Find dy/dx or y' by implicit differentiation. Write the solution of each answers

Question

1. x³y³ = 3xy

Answer: (y-x²y³)/(x³y²-x)

2. x²+2y³+3xy =2x²+y³ +3y²

Answer: (4x-y²-3y)/(2xy+3y²+3x-6y)

Answer 1

1. x³y³ = 3xy

LHS: d(x³y³)/dx = y³[d(x³)/dx] + x³[d(y³)/dx] = y³(3x²) + x³(3y²)(dy/dx) = 3x²y³ + 3x³y²(dy/dx)

RHS: d(3xy)/dx = 3y +3x(dy/dx)

So 3x²y³ + 3x³y²(dy/dx) = 3y + 3x(dy/dx)

Now divide both sides by 3 to get

x²y³ + x³y²(dy/dx) = y + x(dy/dx)

Subtract x²y³ + x(dy/dx) from both sides, then factor the LHS to get

(x³y² – x)(dy/dx) = y – x²y³

Divide both side by (x³y² – x) and our final answer is

dy/dx = (y – x²y³)/(x³y² – x)

which agrees with the answer given in the posted question

2. x²+2y³+3xy = 2x²+y³ +3y²

Subtract 2x²+y³ to get

y³ – x² + 3xy = 3y² (not a necessary step, but it saves a bit of work later)

LHS: d(y³–x²+3xy)/dx = 3y²(dy/dx) – 2x +3y +3x(dy/dx) = (3y² +3x)(dy/dx) + 3y – 2x

RHS: d(3y²)/dx = 6y(dy/dx)

So (3y² +3x)(dy/dx) + 3y – 2x = 6y(dy/dx)

Subtract 3y – 2x + 6y(dy/dx) from both sides, then factor the LHS to get

(3y² + 3x – 6y)(dy/dx) = 2x – 3y

Divide both side by (3y² +3x – 6y) and our final answer is

dy/dx = (2x – 3y)/(3y² + 3x – 6y)

which does not agree with the answer given in the posted question.

Answer 2

x³y³ = 3xy

3x²y³ + x³(3y²)(dy/dx) = 3y + 3x(dy/dx)

(dy/dx)[x³y² - x] = y - x²y³

dy/dx = (y - x²y³) / (x³y² - x)