# Develop an expression for (dy/dx), the slope of the tangent line to the curve for x^2+xy +y^3=2?

### 3 Answers

Relevance

- ComoLv 73 weeks ago
:-

2x + y + x dy/dx + 3y² dy/dx = 0

( x. + 3y² ) dy/dx = - ( y + 2x )

dy/dx = - ( y + 2x ) / ( x + 3y² )

- Ray SLv 73 weeks ago
x² + xy + y³ = 2

2x + (x•y' + y•1) + 3y²•y' = 0

2x + xy' + y + 3y²y' = 0

xy' + 3y²y' = -2x - y

(x + 3y²)y' = -(2x + y)

y' = -(2x + y) / (x + 3y²)

- 3 weeks ago
x^2 + xy + y^3 = 2

2x * dx + x * dy + y * dx + 3y^2 * dy = 0

(2x + y) * dx + (3y^2 + x) * dy = 0

(3y^2 + x) * dy = -(y + 2x) * dx

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

Still have questions? Get your answers by asking now.