We need to find dy/dxdydx.
Start off with differentiating both sides.
d/dx (x^2y+xy^2) = d/dx(3x)ddx(x2y+xy2)=ddx(3x)
Since d/dx(f(x)+g(x)) = d/dx f(x) + d/dx g(x)ddx(f(x)+g(x))=ddxf(x)+ddxg(x) :
d/dx x^2y+ d/dx xy^2 = d/dx 3xddxx2y+ddxxy2=ddx3x
The right hand side is simpler, so let's do that first.
Given any constant cc, d/dx cf(x) = c (d/dx f(x) )ddxcf(x)=c(ddxf(x)), so:
d/dx x^2y+ d/dx xy^2 =3 d/dx xddxx2y+ddxxy2=3ddxx
d/dx x^2y+ d/dx xy^2 =3 (1)ddxx2y+ddxxy2=3(1)
d/dx x^2y+ d/dx xy^2 =3ddxx2y+ddxxy2=3
For the left hand side, we apply the product rule:
d/dx(f(x)g(x)) = f(x)d/dx(g(x)) + g(x)d/dx(f(x))ddx(f(x)g(x))=f(x)ddx(g(x))+g(x)ddx(f(x))
d/dx x^2y + d/dx xy^2 =3ddxx2y+ddxxy2=3
[x^2(d/dx y) + y(d/dx x^2)] + [x(d/dx y^2) + y^2(d/dx x)] =3[x2(ddxy)+y(ddxx2)]+[x(ddxy2)+y2(ddxx)]=3
Differentiating the xx parts is straightforward:
[x^2(d/dx y) + y(2x)] + [x(d/dx y^2) + y^2(1)] =3[x2(ddxy)+y(2x)]+[x(ddxy2)+y2(1)]=3
When differentiating one variable in terms of another, you have to treat the other variable as a function of the one we are differentiating.
For example: d/dx(y) = dy/dx = y'
How this works for more complicated problems uses the chain rule.
For example:
d/dx y^2
d/dx y^2
= 2y d/dx y
= 2y(dy/dx)
A "shortcut" is simply "differentiating as normal, then multiplying a dy/dx (or whatever other variable you're working with.
Going back to the problem:
[x^2(d/dx y) + y(2x)] + [x(d/dx y^2) + y^2(1)] =3
[x^2(dy/dx) + 2xy] + [2xy(dy/dx) + y^2] =3
We need to isolate dy/dx, so we bring all the dy/dx terms to one side, and bring the other terms to the other side:
x^2(dy/dx) + 2xy(dy/dx) = 3 - 2xy - y^2
Now we can factor out dy/dx.
(dy/dx)(x^2+2xy)=3 - 2xy - y^2
dy/dx=(3 - 2xy - y^2)/(x^2+2xy)
