Question

How do you find `(d^2y)/(dx^2)` given `x^2y+xy^2=6`?

Answer 1

`(d^2y)/(dx^2)=(4y(2x+y)(x+y))/(x^2(x+2y)^2)-(2xy^2(2x+y)^2)/(x(x+2y))^3-(2y)/(x(x+2y)`

Explanation:

Generally differentiation problems involve functions i.e. `y=f(x)` - written explicitly as functions of `x`. However, some functions y are written implicitly as functions of `x`.

So what we do is to treat `y` as `y=y(x)` and use chain rule. This means differentiating `y` w.r.t. `y`, but as we have to derive w.r.t. `x`, as per chain rule, we multiply it by `(dy)/(dx)`.

Here we are given `x^2y+xy^2=6`

hence differentiating it we get

`2x xxy+x^2xx(dy)/(dx)+x xx2y(dy)/(dx)+y^2=0`

or `2xy+x^2(dy)/(dx)+2xy(dy)/(dx)+y^2=0` ................(A)

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

Differentiating (A) further, we have

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

or `(d^2y)/(dx^2)(x^2+2xy)+(dy)/(dx)(4x+4y)+2x((dy)/(dx))^2+2y=0`

or `(d^2y)/(dx^2)(x(x+2y))=-4(dy)/(dx)(x+y)-2x((dy)/(dx))^2-2y`

= `4(y(2x+y))/(x(x+2y))(x+y)-2x((y(2x+y))/(x(x+2y)))^2-2y`

and `(d^2y)/(dx^2)=1/(x(x+2y))(4(y(2x+y))/(x(x+2y))(x+y)-2x((y(2x+y))/(x(x+2y)))^2-2y)`

= `(4y(2x+y)(x+y))/(x^2(x+2y)^2)-(2xy^2(2x+y)^2)/(x(x+2y))^3-(2y)/(x(x+2y)`

Answer 2

`(d^2y)/(dx^2) = {6xy(x^2+xy+y^2)(x+y) }/ (x^2+2xy)^3`

Explanation:

There is another (often faster) approach using partial derivatives. Suppose we cannot find `y` explicitly as a function of `x`, only implicitly through the equation `F(x, y) = 0` which defines `y` as a function of `x, y = y(x)`. Therefore we can write `F(x, y) = 0` as `F(x, y(x)) = 0`. Differentiating both sides of this, using the partial chain rule gives us

`(partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))`

And for the second derivative we similarly have:

`dy/dx = G(x,y) => (d^2y)/(dx^2) = (partial G)/(partial x) + (partial F)/(partial y)G(x,y)`

So Let `F(x,y) = x^2y + xy^2 - 6`; Then;

`(partial F)/(partial x) = 2xy+y^2 \ \ \` and `\ \ \ (partial F)/(partial y) = x^2+2xy`

And so:

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

So to get the Second derivative;
Let `G(x,y) =-(2xy+y^2)/(x^2+2xy)`; Then:

`(partial G)/(partial x) = -{(x^2+2xy)(2y) - (2xy+y^2)(2x+2y)} / (x^2+2xy)^2`
`" " = 2y * {2x^2+3xy+y^2 - x^2-2xy} / (x^2+2xy)^2`
`" " = 2y * {x^2+xy+y^2 } / (x^2+2xy)^2`

`(partial G)/(partial y) = -{(x^2+2xy)(2x+2y) - (2xy+y^2)(2x)} / (x^2+2xy)^2`
`" " = 2x*{2xy+y^2-x^2-3xy-2y^2 } / (x^2+2xy)^2`
`" " = -2x*{xy+x^2+y^2 } / (x^2+2xy)^2`

And so using the above formula for the second derivative:

`(d^2y)/(dx^2) = (partial G)/(partial x) + (partial F)/(partial y)G(x,y)`

`" " = {2y(x^2+xy+y^2) } / (x^2+2xy)^2 + { 2x(xy+x^2+y^2)(2xy+y^2) } / {(x^2+2xy)^2(x^2+2xy)}`

Which we can simplify to get:

`(d^2y)/(dx^2) = {2y(x^2+xy+y^2)(x^2+2xy) + 2x(xy+x^2+y^2)(2xy+y^2) } / (x^2+2xy)^3`

`" " = {2xy(x^2+xy+y^2){ (x+2y) + (2x+y) } }/ (x^2+2xy)^3`

`" " = {6xy(x^2+xy+y^2)(x+y) }/ (x^2+2xy)^3`