Question

How do you find the second derivative of `y^2 + x + sin y = 9`?

Answer 1

The second derivative is `y''=(sin(y)-2)/(2y+cos(y))^3`

Explanation:

Implicit differentiation gives us

`d/dx(y^2)+d/dx(x)+d/dx(sin(y))=d/dx(9)`
`2y dy/dx+1+cos(y) dy/dx=0`
`2y dy/dx+cos(y) dy/dx=-1`
`dy/dx(2y +cos(y))=-1`. Therefore,

`y'=dy/dx=(-1)/(2y +cos(y))`. This implies that `2y+cos(y)!=0`.

Taking the second implicit derivative gives us
`d/dx[dy/dx(2y +cos(y))=-1]`

By application of the Product Rule, we have
`d/dx(dy/dx)(2y +cos(y))+dy/dx(2dy/dx-sin(y) dy/dx)=0`
`(d^2y)/dx^2(2y +cos(y))+2((dy)/dx)^2-sin(y) ((dy)/dx)^2=0`
`(d^2y)/dx^2(2y +cos(y))=sin(y) ((dy)/dx)^2-2((dy)/dx)^2`
`(d^2y)/dx^2=(sin(y) ((dy)/dx)^2-2((dy)/dx)^2)/((2y +cos(y)))` or more succinctly

`y''=(y')^2(sin(y)-2)/(2y+cos(y))`
`y''=((-1)/(2y +cos(y)))^2(sin(y)-2)/(2y+cos(y))`
`y''=1/(2y +cos(y))^2(sin(y)-2)/(2y+cos(y))`
`y''=(sin(y)-2)/(2y+cos(y))^3`

Answer 2

I get: `y'' = (-2+siny)/(2y+cosy)^3`

Explanation:

`x+y^2+siny = 9`

`d/dx(x+y^2+siny) = d/dx(9)`

`1 + 2y dy/dx + cosy dy/dx = 0`

`dy/dx = (-1)/(2y+cosy) = -(2y+cosy)^-1`

`d/dx(dy/dx) = 1(2y+cosy)^-2*[d/dx(2y+cosy)]`

`= (2y+cosy)^-2 [2 dy/dx - siny dy/dx]`

`= (2y+cosy)^-2 [2 - siny] dy/dx`

`= (2y+cosy)^-2 [2 - siny] [-(2y+cosy)^-1]`

`= -(2-siny)(2y+cosy)^-3`

`= (-2+siny)/(2y+cosy)^3`