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)
Find y' :
y= (8-2x^2)/(root of (8-x^2))
1 answer