How do you implicitly differentiate xy + 3x + 5x^2 = 4/y^3-xy+1?

1 Answer

y'=(-10x-2y-3)/(2x+12/y^4)

Explanation:

The given equation

xy+3x+5x^2=4/y^3-xy+1

Differentiate both sides of the equation with respect to x

d/dx(xy+3x+5x^2)=d/dx(4/y^3-xy+1)

d/dx(xy)+d/dx(3x)+d/dx(5x^2)=d/dx(4/y^3)-d/dx(xy)+d/dx(1)

xy'+y(1)+3+10x=4(-3)y^(-3-1)*y'-[xy'+y*1]+0

simplify

xy'+y+3+10x=-12y^(-4)*y'-xy'-y

transpose all terms containing y' on one side of the equation

xy'+12y^(-4)y'+xy'=-y-y-10x-3

2xy'+12y^(-4)y'=-2y-10x-3

(2x+12y^(-4))y'=-2y-10x-3

y'=(-10x-2y-3)/(2x+12/y^4)

God bless...I hope the explanation is useful.