Question

Find y' :

y= (8-2x^2)/(root of (8-x^2))

Answer 1

using the product rule

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

= -(8-x^2)^(-1/2)[8-x^2 + 4x]

= (8-x^2)^(-1/2)(x^2 - 4x - 8)