Let, #y=F(x)=sqrt{(x-8)/(x^2-2)}={(x-8)/(x^2-2)}^(1/2).#
#:. lny=1/2ln((x-8)/(x^2-2))=1/2{ln(x-8)-ln(x^2-2)}.#
#:. d/dx{lny}=1/2d/dx{ln(x-8)-ln(x^2-2)}.#
By the Chain Rule, then, we have,
#:. d/dy(lny)*dy/dx#
#=1/2[1/(x-8)*d/dx(x-8)-1/(x^2-2)*d/dx(x^2-2)}.#
#:. 1/y*dy/dx=1/2{1/(x-8)*1-1/(x^2-2)*2x},#
#=1/2{{(x^2-2)-2x(x-8)}/{(x-8)(x^2-2)}},#
#=1/2{(-x^2+16x-2)/{(x-8)(x^2-2)}}.#
# rArr dy/dx=-(y(x^2-16x+2))/{2(x-8)(x^2-2)}.#
But, #y={(x-8)/(x^2-2)}^(1/2).#
#:. dy/dx=-{(x-8)/(x^2-2)}^(1/2)*(x^2-16x+2)/{2(x-8)(x^2-2)}, i.e., #
# dy/dx=F'(x)=-((x^2-16x+2))/{2(x-8)^(1/2)(x^2-2)^(3/2)}.#
Enjoy Mayhs.!