find y' implicit differentiation x^3-y/1-y^3=x

Answer 1

You would have to use the quotient rule here. The top function is x³-y and the bottom function is 1-y³

From the chain rule the left hand side becomes [ (1-y³) • d/dx(x³-y) - (x³-y) • d/dx(1-y³) ] / (1-y³)²

d/dx (x³ - y) = 3x² - dy/dx = 3x² - y'

d/dx (1-y³) = -3y²y'

Now we plug in to get:

[ (1-y³) • (3x² -y') - (x³-y) • (-3y²y' ) ] / (1-y³)²

The derivative of the right hand side is just 1 so,

[ (1-y³) • (3x² -y') - (x³-y) • (-3y²y' ) ] / (1-y³)² = 1

(1-y³) • (3x² -y') - (x³-y) • (-3y²y' ) = (1-y³)²

Split the first product

(1-y³)(3x²) - (1-y³)y' - (x³-y)(-3y²y' ) = (1-y³)²

Simplify

(1-y³)(3x²) - (1-y³)y' + 3y²(x³-y)y' = (1-y³)²

Put everything multiplied by y' on one side and everything not multiplied by y' on the other

- (1-y³)y' + 3y²(x³-y)y' = (1-y³)² - (1-y³)(3x²)

Combine the y' terms

y' [ 3y²(x³-y) - (1-y³)] = (1-y³)² - (1-y³)(3x²)

Solve for y' by dividing

y' = [ (1-y³)² - (1-y³)(3x²) ] / [ 3y²(x³-y) - (1-y³)]

Answer 2

I am assuming you mean (x^3-y)/(1-y^3)=1

(1-y^3)(3x^2 - y') - (x^3-y)(-3y^2 y') = 0

(1-y^3)(3x^2) - (1-y^3)y' + (3y^2)(x^3-y) + 3y^3 y' = 0

[(3y^3)-[1-y^3]}y' = (3y^2)(x^3-y) - (y^3-1)(3x^2)

[ 3y^3 - 1 + y^3] y' = (3y^2)(x^3-y) - (y^3-1)(3x^2)

y' = [(3y^2)(x^3-y) - (y^3-1)(3x^2)] / (4y^3-1)