Question

Question #e1184

Answer 1

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

Explanation:

`2x^3y+3xy^2=5`

First we take `d/dx` of every term.

`d/dx[2x^3y]+d/dx[3xy^2]=d/dx[5]`

Then we use the product rule, `d/dx[f(x)g(y)]=f(x)d/dx[g(y)]+g(y)d/dx[f(x)]`

`yd/dx[2x^3]+2x^3d/dx[y]+3xd/dx[y^2]+y^2d/dx[3x]=d/dx[5]`

`y(6x^2)+2x^3d/dx[y]+3xd/dx[y^2]+y^2(3)=0`

the chain rule tells us that `d/dx=d/dyxx(dy)/(dx)`

`y(6x^2)+(dy)/(dx)2x^3d/dy[y]+(dy)/(dx)3xd/dy[y^2]+y^2(3)=0`

`6x^2y+(dy)/(dx)2x^3(1)+(dy)/(dx)3x(2y)+3y^2=0`

`6x^2y+(dy)/(dx)2x^3+(dy)/(dx)6xy+3y^2=0`

Now we rearrange:
`(dy)/(dx)2x^3+(dy)/(dx)6xy=-3y^2-6x^2y`

`(dy)/(dx)(2x^3+6xy)=-3y^2-6x^2y`

`(dy)/(dx)=(-3y^2-6x^2y)/(2x^3+6xy)`

`color(white)((dy)/(dx))=-(3y^2+6x^2y)/(2x^3+6xy)`

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