And another... find dy/dx

[Marcia]

Find dy/dx of y=Square root of (x^2 -2) / x^3 at x=Square root of (3).

dy/dx = d/dx = x^3(1/2(x^2 - 2)^(-1/2) - 3x(Square root of (x^2 -2)

Am I starting this right?

What do i do now?

 

[Unco]

Am I starting this right?

Yep! Provided you meant

dy/dx = x^3(1/2(x^2 - 2)^(-1/2) - (3x^2)(Square root of (x^2 -2) (1))

Simplify x^3(1/2(x^2 - 2)^(-1/2), and then subsitute x = sqrt(3) into (1).