Arc Length of curve 24xy = x^4 + 48 from x = 2 to x = 4

The steps are below:

My question is,
where is the "-1/2" coming from in the "1 + (dy/dx)^2" step?

Thx
y = x^3/24 + 2x^(-1).

dy/dx = x^2/8 - 2x^(-2)

So, 1 + (dy/dx)^2
= 1 + (x^4/64 - 1/2 + 4x^(-4))
= x^4/64 + 1/2 + 4x^(-4)
= (x^2/8 + 2x^(-2))^2

L =
∫(2 to 4) sqrt[(x^2/8 + 2x^(-2))^2] dx
= ∫(2 to 4) (x^2/8 + 2x^(-2)) dx
= (x^3/24 - 2x^(-1)) {for x = 2 to 4}
= 17/6

It comes from (dy/dx)^2

dy/dx = x^2/8 - 2x^(-2)

so

(dy/dx)^2 = { x^2 / 8 - 2x^{-2} }^2

= x^4 / 64 - (x^2 / 8)(2 / x^2) - (x^2 / 8)(2 / x^2) + 4 / x^4

= x^4 / 64 - 2/8 - 2/8 + 4 / x^4

= x^4 - 1/4 - 1/4 + 4 / x^4

(dy/dx)^2 = x^4 - 1/2 + 4 / x^4

so,

1 + (dy/dx)^2 = 1 + (x^4/64 - 1/2 + 4x^(-4))