Question

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

Answer 1

`d^(2y)/(dx^2) = -4/(5y^3)`

Explanation:

The equation does not express `y` explicitly in terms of `x` as in `y=f(x)`, instead we have `g(y)=f(x)`, so when we differentiate we apply the chain rule so that we differentiate `g(y)` wrt `y` rather than `x`, as in:

`g(y) = f(x)`
`:. d/dx g(y) = d/dx f(x)`
`:. dy/dx d/dy g(y) = f'(x)`
`:. g'(y) dy/dx= f'(x)`

So for `5x^2 = 5y^2 + 4` we have:

`10x = 10y dy/dx`
`y dy/dx = x` ..... [1]

To find the second derivative we need to differentiate a second time again implicitly and using the product rule:

`y dy/dx = x`
`:. (y)(d/dxdy/dx) + (d/dxy)(dy/dx) = 1`
`:. y(d^(2y)/(dx^2)) + (dy/dx)^2 = 1`

From [1] we have `dy/dx=x/y` so substituting gives:

`:. y(d^(2y)/(dx^2)) + (x/y)^2 = 1`
`:. yd^(2y)/(dx^2) = 1 - x^2/y^2`
`:. yd^(2y)/(dx^2) =(y^2-x^2)/y^2`

so we have

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

But from the original equation we have;
`5x^2 = 5y^2 + 4 => 5(y^2 - x^2) = -4 => y^2 - x^2 = -4/5`

Hence,

`d^(2y)/(dx^2) =(-4/5)/y^3 = -4/(5y^3)`