How do you find (dy)/(dx)dydx given -3x^2y^2-2y^3+5=5x^23x2y22y3+5=5x2?

1 Answer
Sep 22, 2017

(dy)/(dx)=(-(5x+3xy^2))/(3y(x^2+y)dydx=(5x+3xy2)3y(x2+y)

Explanation:

we want

d/(dx)(-3x^2y^2)-d/(dx)(2y^3)+d/(dx)(5)=d/(dx)(5x^2)ddx(3x2y2)ddx(2y3)+ddx(5)=ddx(5x2)

We will differentiate implicitly.

The first term will also need the product rule

color(red)(d/(dx)(uv)=v(du)/(dx)+u(dv)/(dx))ddx(uv)=vdudx+udvdx

y^2(-6x)+(-3x^2)2y(dy)/(dx)-6y^2(dy)/(dx)+0=10xy2(6x)+(3x2)2ydydx6y2dydx+0=10x

-6xy^2-6x^2y(dy)/(dx)-6y^2(dy)/(dx)=10x6xy26x2ydydx6y2dydx=10x

now rearrange for (dy)/(dx)dydx and tidy up.

(dy)/(dx)(-6x^2y-6y^2)=10x+6xy^2dydx(6x2y6y2)=10x+6xy2

(dy)/(dx)=(10x+6xy^2)/(-6y(x^2+y)dydx=10x+6xy26y(x2+y)

(dy)/(dx)=(-(5x+3xy^2))/(3y(x^2+y)dydx=(5x+3xy2)3y(x2+y)