find dy/dx given that xy^2-x^2y+x^3=4

Question

tommo_lcfc · Answer

This is an example of implicit differentiation. You need to use the product rule to differentiate the first two terms and then add a dy/dx to the terms you are differentiating with respect to y.

Differentiating gives
$$2xy dy/dx + y^2 - 2xy + x^2 dy/dx + 3x^2 = 0$$

If we put all the terms with dy/dx to the other side, we get
$$2xy - y^2 = (2xy + x^2) dy/dx$$

Therefore $$dy/dx = (2xy - y^2) / (2xy + x^2)$$

There is a common factor in both terms so this can be re-written as
$$dy/dx = y(2x+y)/x(2y+x)$$

tommo_lcfc · Answer

Sorry, sign error, the + x^2 should be a -x^2 so we get
dy/dx = y(2x-y)/x(2y-x)

anonymous · Answer

thanks

anonymous · Answer

xy^2            -     x^2   y            +    x^3         =   4
( 2xy *y +y^2) -   (x^2 *y' + 2xy)  + 3x^2           =   0
 3x^2 + 2xy        + y^2 +    2xy *y' +   x^2 y'     = 0
 (x^2 + 2xy) y'             = - ( 3x^2 + 2xy + y^2 )
=> y' =  - ( 3x^2 + 2xy + y^2 ) / (x^2 + 2xy)