text: I'm not sure if you meant the fn to be\nf(X) = (x+4)(x^2+4) /(x-5) or\nf(X) = {(x+4)(x^2+4)/x} -5 \n----\nIf case 1: Find the derivative of the function\nf(X) = (x+4)(x^2+4) /(x-5)\n\nLet g(X)= (X+4)(x^2+4)\ng'(X)= (x^2+4) + (x+4)2x\n=x^2 + 4 +2x^2 +8x\n=3x^2+8x+4\n\nf(x)= g(x)/(x-5)=g(x)*(x-5)^-1\nf'(x)=g'(x)*(x-5)^-1 + g(x)*(-1)(x-5)^-2\n\nf'(x)=[g'(x)(x-5)- g(x)]/[(x-5)^2]\n\nf'(x)={(3x^2+8x+4)(x-5)-(X+4)(x^2+4) }/[(x-5)^2]\n={3x^3+8x^2+4x-15x^2-40x-20-x^3-4x^2-4x-16}/[(x-5)^2]\n={2x^3-11x^2-40x-36}/[(x-5)^2]\n\n\nI 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. \n\nCase 2:\nIf it is f(X) = (x+4)(x^2+4) / (x-5) \nthen above answer should be correct if there where no errors in the steps. But if the eq is:\nf(X) = {(x+4)(x^2+4)/x} -5 then the answer is different.\nlet g(x)= {(x+4)(x^2+4)/x}\n=> f(x)=g(x)-5\nf'(x)=g'(x) since d(-5)/dx  =0\n\ng'= (-1)(x+4)(x^2+4)/x^2 +\n{(x^2+4)+ (x+4)2x}/x\n={-(x+4)(x^2+4)+x(x^2+4)+2(x^2)(x+4)}/x^2\n={(x^2+4)(x-x-4)+ 2(x^2)(x+4)}/x^2\n={-4x^2-16 + 2x^3+8x^2}/x^2\n={2x^3 + 4x^2 -16}/x^2\n=2{x^3+2x^2-8}/x^2\n\n=> f'(x)= 2{x^3+2x^2-8}/x^2\n\nCase 1:\nf'(x)=\n{2x^3-11x^2-40x-36}/(x-5)^2\nwhere\nf(X) = (x+4)(x^2+4) /(x-5)\n\nCase 2:\nf'(x)= 2{x^3+2x^2-8}/x^2\nwhere\nf(x) = {(x+4)(x^2+4)/x} -5