text: find the derivative of the function.?   i'm not sure if you meant the fn to be f(x) = (x+4)(x^2+4) /(x-5) or f(x) = {(x+4)(x^2+4)/x} -5  ---- if case 1: find the derivative of the function f(x) = (x+4)(x^2+4) /(x-5)  let g(x)= (x+4)(x^2+4) g'(x)= (x^2+4) + (x+4)2x =x^2 + 4 +2x^2 +8x =3x^2+8x+4  f(x)= g(x)/(x-5)=g(x)*(x-5)^-1 f'(x)=g'(x)*(x-5)^-1 + g(x)*(-1)(x-5)^-2  f'(x)=[g'(x)(x-5)- g(x)]/[(x-5)^2]  f'(x)={(3x^2+8x+4)(x-5)-(x+4)(x^2+4) }/[(x-5)^2] ={3x^3+8x^2+4x-15x^2-40x-20-x^3-4x^2-4x-16}/[(x-5)^2] ={2x^3-11x^2-40x-36}/[(x-5)^2]   i just did a hand check it with pen & paper & i edited since i found a mistake in my initial calc. using the keyboard is my biggest challenge. now, i might have misread equ., so i did two cases, depending which is the eq. that you wanted the derivative of.   case 2: if it is f(x) = (x+4)(x^2+4) / (x-5)  then above answer should be correct if there where no errors in the steps. but if the eq is: f(x) = {(x+4)(x^2+4)/x} -5 then the answer is different. let g(x)= {(x+4)(x^2+4)/x} => f(x)=g(x)-5 f'(x)=g'(x) since d(-5)/dx  =0  g'= (-1)(x+4)(x^2+4)/x^2 + {(x^2+4)+ (x+4)2x}/x ={-(x+4)(x^2+4)+x(x^2+4)+2(x^2)(x+4)}/x^2 ={(x^2+4)(x-x-4)+ 2(x^2)(x+4)}/x^2 ={-4x^2-16 + 2x^3+8x^2}/x^2 ={2x^3 + 4x^2 -16}/x^2 =2{x^3+2x^2-8}/x^2  => f'(x)= 2{x^3+2x^2-8}/x^2  case 1: f'(x)= {2x^3-11x^2-40x-36}/(x-5)^2 where f(x) = (x+4)(x^2+4) /(x-5)  case 2: f'(x)= 2{x^3+2x^2-8}/x^2 where f(x) = {(x+4)(x^2+4)/x} -5