Question

How do you find ((d^2)y)/(dx^2)?

Question: Find (d^y)/(dx^2) by implicit differentiation when 2xy+2y^2=13

Answer 1

`y''=pm13/(26 + x^2)^(3/2)`

Explanation:

Defining

`f(x,y(x))=2xy(x)+2y(x)^2-13=0` then

`(df)/(dx)=2y+2xy'+4yy'=0` and

`d/(dx)(df)/(dx)=4y'+2xy''+4y'^2+4yy''=0`

after substitution of `y'` we have

`y''=(2xy+2y^2)/(x+2y)^3=13/(x+2y)^3`

but `y =1/2 (-x pm sqrt[26 + x^2])` so finally

`y''=pm13/(26 + x^2)^(3/2)`

Answer 2

`(d^2y)/(dx^2) = 13/(x+2y)^3`

Explanation:

`2xy+2y^2=13`

Differentiating wrt `x` and applying the product rule gives us:

`2{ (x)(dy/dx) + (1)(y) } +4ydy/dx = 0`
`xdy/dx + y +2ydy/dx = 0 => dy/dx = -y/(x+2y)`

Differentiating again wrt `x` and applying the product rule (twice) gives us:

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

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

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

Putting over a common denominator gives:

`(d^2y)/(dx^2) = ((2y)(x+2y) - (2y^2))/(x+2y)^3`
`:. (d^2y)/(dx^2) = ((2xy+4y^2-2y^2))/(x+2y)^3`
`:. (d^2y)/(dx^2) = ((2xy+2y^2))/(x+2y)^3`
`:. (d^2y)/(dx^2) = 13/(x+2y)^3` (as `2xy+2y^2=13`)