The quotient rule; given #f(x)!=0#
if #h(x)= f(x)/g(x)# ; then #h'(x)= [g(x)*f'(x) -f(x)*g'(x)]/(g(x))^2#
given #h(x)= (x^2+x+3)/root()(x-3)#
let #f(x)= x^2 +x+3#
#color (red) (f'(x)= 2x+1)#
let #g(x)= root()(x-3) = (x-3)^(1/2)#
#color(blue)(g'(x)= 1/2 (x-3)^(1/2-1)= 1/2 (x-3)^(-1/2)#
#h'(x)= [(x-3)^(1/2)*color(red)((2x+1))-color(blue)(1/2(x-3)^(-1/2))(x^2 +x+3)]/(root()[(x-3)]^2#
Factor out the greatest common factor #1/2 (x-3)^(-1/2)#
#h'(x)= 1/2 (x-3)^(-1/2)[(x-3)(2x+1)-(x^2 +x+3)]/(x-3)#
#=>h'(x) = 1/2[(x^2 +x-6x-3-x^2 -x-3)]/(x-3)^(3/2)#
#h'(x)= (-6x-6)/(2(x-3)^(3/2))#
#h'(x)= -[6(x+1)]/(2(x-3)^(3/2))#
#color(red)(h'(x)= -[3(x+1)]/((x-3)^(3/2)))# Answer
