Question

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

Answer 1

`(d^2y)/dx^2 = -1/y^3`

Explanation:

If an equation does not express `y` explicitly in terms of `x` as in `y=f(x)`, and instead we have `g(y)=f(x)`, then 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)`

This will typically result in `dy/dx` being a function of `x` and `y`, rather than `dy/dx` being a function of `x` alone as we usually get

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

`d/dx(2x) + d/dx(y^2) = d/dx(5)`
`:. 2 + dy/dxd/dy(y^2) = 0)`
`:. 2 + dy/dx2y = 0`
`:. 1 + dy/dxy = 0`
`:. dy/dx = -1/y` ..... [1]

Differentiating again wrt `x` we get:

`(d^2y)/dx^2 = d/dx(-1/y)`
`:. (d^2y)/dx^2 = -dy/dxd/dy(1/y)`
`:. (d^2y)/dx^2 = -dy/dx(-y^-2)`
`:. (d^2y)/dx^2 = dy/dx(1/y^2)`

And substituting from [1] we get:

`(d^2y)/dx^2 = (-1/y)(1/y^2)`
`:. (d^2y)/dx^2 = -1/y^3`