Question

How do you find `(d^2y)/(dx^2)` for `x^2+4y^2=5`?

Answer 1

`(d^2y)/(dx^2)[x^2+4y^2=5] = (-5)/(16y^3)`

Explanation:

First find `(dy)/(dx)` by implicitly differentiating `x^2+4y^2=5`:
`x^2+4y^2=5`
`2x+8y((dy)/(dx))=0`
`((dy)/(dx))(8y) = -2x`
`(dy)/(dx) = (-2x)/(8y)`
`(dy)/(dx) = (-x)/(4y)`

Now implicitly differentiate `(dy)/(dx)`:

Use the quotient rule:
`(d^2y)/(dx^2) = [(4y)(-1)-(-x)(4(dy/dx))]/((4y)^2)`

Simplify:
`(d^2y)/(dx^2) = [-4y+4x(dy/dx)]/(16y^2)`

Substitute `(dy)/(dx) =(-x)/(4y)`

`(d^2y)/(dx^2)= [-cancel(4)y+cancel(4)x((-x)/(4y))]/(4cancel(16)y^2)`

Simplify:
`(d^2y)/(dx^2)= [(-y)/(4y^2)]-[(x^2)/((4y)(4y^2))]`

`(d^2y)/(dx^2)= [(-1)/(4y)]-[(x^2)/(16y^3)]`

Make a common denominator to combine into one fraction:

`(d^2y)/(dx^2)= (-4y^2-x^2)/(16y^3)`
`(d^2y)/(dx^2)= (-(4y^2+x^2))/(16y^3)`

Recall that the original equation states that `4y^2+x^2=5`, so:

`(d^2y)/(dx^2)= (-5)/(16y^3)`