Question

Find dy/dx for the following:

1) x^3 + 3x^2*y + y^3 = 4
I get -3(x+2y)/x^2 + y^2

it's wrong though.

2) x + y^2= xy^2
I get y^2 + 1/ 2y- 2xy
this is also wrong.

can you explain thanks

Answer 1

x^3 + 3x^2*y + y^3 = 4 --------->

d[x^3 + 3x^2*y + y^3] = d(4) = 0 --->

3 x^2 dx + 6xy dy + 3x^2 dy + 3y^2 dy = 0
--->

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

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

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

-[x/(x+y)]^2

Answer 2

Hmmm I get

x^3 + 3x^2*y + y^3 = 4
3x^2 + 6xy + 3x^2 y' + 3y^2 y' = 0
y'(3x^2 + 3y^2) = -3x^2-6xy
y' = -x(x+2y)/(x^2+y^2)
you must have lost a factor of 3 somewhere


x + y^2= xy^2
1 + 2yy' = y^2 + 2xyy'
y'(2y-2xy) = y^2-1
y' = (y^2-1) / 2y(1-x)
you must have missed a sign change somewhere