Question

How do you find `(d^2y)/(dx^2)` given `2x^2-3y^2=4`?

Answer 1

`(d^2y)/(dx^2) = -8/(9y^3)`

Explanation:

`2x^2-3y^2=4`

Differentiating implicitly we get;
`4x-6ydy/dx=0`
`:. dy/dx=(4x)/(6y)`
`:. dy/dx=(2x)/(3y)`

So, using the quotient rule
`d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2`
the second derivative is given by:

`(d^2y)/(dx^2) = ((3y)(d/dx2x)-(2x)(d/dx3y))/(3y)^2`
`:. (d^2y)/(dx^2) = (6y-(2x)(dy/dxd/dy3y))/(9y^2)`
`:. (d^2y)/(dx^2) = (6y-(2x)(3dy/dx))/(9y^2)`
`:. (d^2y)/(dx^2) = (6y-6xdy/dx)/(9y^2)`
`:. (d^2y)/(dx^2) = (6y-6x((2x)/(3y)))/(9y^2)`
`:. (d^2y)/(dx^2) = (6y-4x^2/y)/(9y^2)`
`:. (d^2y)/(dx^2) = (6y^2-4x^2)/(9y^3)`
`:. (d^2y)/(dx^2) = 2(3y^2-2x^2)/(9y^3)`

But `2x^2-3y^2=4 => 3y^2-2x^2=-4`
`:. (d^2y)/(dx^2) = 2(-4)/(9y^3)`
`:. (d^2y)/(dx^2) = -8/(9y^3)`

Answer 2

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

Explanation:

Computing the second implicit differentiation is determined by computing the first implicit derivative .

`(2dx^2)/dx-(3dy^2)/dx=(d4)/dx`
`rArr2(2x)-3xx2yxx(dy)/dx=0`
`rArr4x-6yxx(dy)/dx=0`
`rArr-6yxx(dy)/dx=-4x`
`rArr(dy)/dx=(-4x)/(-6y)`

`rArrcolor(blue)((dy)/dx=(2x)/(3y))`

`color(red)((d^2y)/dx^2=??)`

`color(red)((d^2y)/dx^2)=((d(2x))/(dx)xx3y-3((dy)/dx)xx2x)/(3y)^2`

`color(red)((d^2y)/dx^2)=(2xx3y-6x((dy)/dx))/(3y)^2`

`color(red)((d^2y)/dx^2)=(6y-6x((dy)/dx))/(9y^2)`

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