Question

`x^2y + xy = 6`. Find `dx/dy`?

Answer 1

`dx/dy=-(x^2(x+1)^2)/(6(2x+1))`.

Explanation:

`x^2y+xy=6`

Treating `x` as a function of `y`, we get:

`=>d/dy (x^2y)+d/dy (xy)=d/dy(6)`
`=>[(d(x^2))/dy*y+x^2* (d(y))/dy]+[(d(x))/dy*y + x*(d(y))/dy]=0`

Let `x'=dx/dy`.

`=>2x x'y+x^2(1)+x'y+x(1)=0`
`=>2x x'y+x'y+x^2+x=0`
`=>x'(2xy+y)="-"(x^2+x)`
`=>x'=-(x^2+x)/(2xy+y)=-(x(x+1))/(y(2x+1))`

Also:
`x^2y+xy=6`

`=>y(x^2+x)=6`
`=>y=6(x^2+x)^"-1"=6/(x(x+1))`

Substituting this into our equation for `x'` gives us

`x'=-(x(x+1))/([6/(x(x+1))] (2x+1))`

`color(white)(x')=-(x^2(x+1)^2)/(6(2x+1))`

Another way:

We could also recognize that, since `y=6/(x(x+1))`, we have

`dy/dx=6("-"1)(x^2+x)^"-2"(2x+1)=-(6(2x+1))/((x^2+x)^2)`

And reciprocating this gives

`dx/dy=-(x^2+x)^2/(6(2x+1))=-([x(x+1)]^2)/(6(2x+1))=-(x^2(x+1)^2)/(6(2x+1))`

as before.

Answer 2

I got:

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

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A nice trick is to recognize that `(dy)/(dx) = 1/((dx)/(dy))`. Let's try both ways (finding `(dy)/(dx)` and taking the reciprocal, vs. finding `(dx)/(dy)`).

`f(x,y) = x^2y + xy = 6`

From the chain rule, `(df)/(dx) = (df)/(dy)(dy)/(dx)`, so, using the product rule twice:

`=> (x^2(d)/(dy) [y]cdot(dy)/(dx) + (d)/(dx) [x^2]y) + (x(d)/(dy) [y]cdot(dy)/(dx) + y(d)/(dx) [x]) = 0`

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

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

`=> (dy)/(dx)[x^2 + x] = -(2xy + y)`

`=> (dy)/(dx) = -(2xy + y)/[x^2 + x]`

So, `color(blue)((dx)/(dy) = -(x^2+x)/(2xy + y))`.

Let's check the other way. If `x = x(y)`, then, using the product rule twice:

`=> (x^2 (d)/(dy) [y] + y d/(dx) [x^2]cdot(dx)/(dy)) + (x(d)/(dy) [y] + y(d)/(dx) [x]cdot(dx)/(dy)) = 0`

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

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

`=> (dx)/(dy)[2xy + y] = -(x + x^2)`

`=> color(blue)((dx)/(dy) = -(x + x^2)/[2xy + y])`

Indeed, same result both ways.