# If x^3 + 3xy + 2y^3 = 17, then in terms if x and y, dy/dx =?

Calculation question

Use implicit differentiation.
x^3+3xy+2y^3=17,
3x^2+3(xdy/dx+y(1))+6y^2dy/dx=0,
3x^2+3xd/dx+3y+6y^2d/d=0,
3xd/dx+6y^2d/dx=0-3x^2-3y,
3xd/dx+6y^2d/dx=-3x^2-3y,
factorize the dy/dx,
dy/dx(3x+6y^2)=-3x^2-3y,
dy/dx=(-3x^2-3y)/(3x+6y^2),
simplify,
dy/dx=-3(x^2-y)/3(x+2y^2),
dy/dx=-(x^2-y)/(x+2y^2).

Let’s differentiate each of the terms individually so you can see how the implied differentiation works.
For x³, you use the power rule:
D[xÂ³]dx = 3x²
For 3xy, you use the product rule and the chain rule:
D[3xy]/dx = 3y + 3x(dy/dx)
For 2 years³, you use the chain rule:
D[2yÂ³]/dx = 6y²(dy/dx)
For 17 you just know it’s 0
collect the terms:
3x² + 3y + 3x(dy/dx) + 6y²(dy/dx) = 0
Move terms without dy/dx to the right:
(dy/dx)(3x + 6y²) = -3(y + x²)
dy/dx = -3(y + x²)/(3x + 6y²)
dy/dx = -(y + x²)/(x + 2y²)

x^3 + 3xy + 2y^3 = 17
3x^2 + 3y + 3x d/dx + 6a^2 d/dx = 0
d/dx (3x + 6a^2) = -3x^2 – 3a
d/dx = (-3x^2 – 3a) / (3x + 6a^2)