Question

What the second derivative? `x^(2/3)+ y^(2/3) = 1` Help me please

Answer 1

see below

Explanation:

`x^(2/3)+y^(2/3) = 1`

Differentiate implicitly to get

`2/3x^(-1/3) + 2/3y^(-1/3) dy/dx = 0`

so, (muliply through by `3/2`)

`x^(-1/3) + y^(-1/3) dy/dx = 0`

`y^(-1/3) dy/dx = -x^(-1/3)`

`dy/dx = -x^(-1/3)/y^(-1/3)`

`dy/dx = -y^(1/3)/x^(1/3)`

Differentiate again.

`(d^2y)/dx^2 = -(x^(1/3)(1/3)y^(-2/3)(dy/dx) - y^(1/3)(1/3)x^(-2/3))/(x^(2/3)`

Factor out 1/3 and replace `dy/dx`

`= -1/3((x^(1/3)y^(-2/3)(-y^(1/3)/x^(1/3))-x^(-2/3)y^(1/3)))/x^(2/3)` `" "` simplify

`= 1/3((y^(-2/3)y^(1/3)+x^(-2/3)y^(1/3)))/(x^(2/3))` `" "` factor out `y^(1/3)`

`= 1/3y^(1/3)((y^(-2/3)+x^(-2/3))/x^(2/3))` * `(x^(2/3)y^(2/3))/(x^(2/3)y^(2/3))`

`= 1/3y^(1/3)((x^(2/3)+y^(2/3))/(x^(4/3)y^(2/3)))`

`= ((x^(2/3)+y^(2/3))/(3x^(4/3)y^(1/3)))`

Now recall that `x^(2/3)+y^(2/3)=1` so we finish with

`(d^2y)/dx^2 = 1/(3x^(4/3)y^(1/3))`

Answer 2

Given that, `x^(2/3)+y^(2/3)=1.............<<0>>.`

Diff.ing both sides w.r.t. `x,` we have,

`d/dx(x^(2/3))+d/dx(y^(2/3))=d/dx(1).`

`:. 2/3x^(2/3-1)+d/dy(y^(2/3))*d/dx(y)=0.......[because," the Chain Rule],"`

`:. 2/3x^(-1/3)+2/3y^(-1/3)*dy/dx=0.`

`:. x^(-1/3)+y^(-1/3)*dy/dx=0........<<1>>.`

`:. dy/dx=-x^(-1/3)/y^(-1/3)......<<2>>.`

Rediff.ing `<<1>>" w.r.t. "x,`

`-1/3x^(-1/3-1)+y^(-1/3)*d/dx{dy/dx}+dy/dx*d/dx(y^(-1/3))=0.`

`:. -1/3x^(-4/3)+y^(-1/3)*(d^2y)/dx^2+{-1/3y^(-4/3)*dy/dx}dy/dx=0.`

`:. -1/3x^(-4/3)+y^(-1/3)(d^2y)/dx^2-1/3y^(-4/3)(dy/dx)^2=0.`

`:. y^(-1/3)(d^2y)/dx^2=1/3{x^(-4/3)+y^(-4/3)(dy/dx)^2},`

`=1/3{x^(-4/3)+y^(-4/3)(-x^(-1/3)/y^(-1/3))^2}...[because, <<2>>],`

`=1/3{x^(-4/3)+y^(-4/3)(x^(-2/3)/y^(-2/3))},`

`=1/3{(x^(-4/3)y^(-2/3)+y^(-4/3)x^(-2/3)}/y^(-2/3)},`

`=1/3{(x^(-2/3)y^(-2/3))(x^(-2/3)+y^(-2/3))}/y^(-2/3),`

`=1/3x^(-2/3)(x^(-2/3)+y^(-2/3)),`

`=1/3x^(-2/3)(1/x^(2/3)+1/y^(2/3)),`

`=1/3x^(-2/3)(x^(2/3)+y^(2/3))/(x^(2/3)y^(2/3)),`

`=1/3x^(-2/3)(x^(-2/3)y^(-2/3))...[because, <<0>>].`

`:. (d^2y)/dx^2=1/y^(-1/3){1/3x^(-4/3)y^(-2/3)}, i.e.,`

`(d^2y)/dx^2=1/3x^(-4/3)y^(-1/3).`