[tex]h(x)=\dfrac{x^2}{8-8x}\\\\h'(x)=\left(\dfrac{x^2}{8-8x}\right)'\\\\\text{use}\ \left(\dfrac{f(x)}{g(x)}\right'=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\\\\f(x)=x^2\to f'(x)=(x^2)'=2x\\\\g(x)=8-8x\to g'(x)=(8-8x)'=-8\\\\\text{substitute}\\\\h'(x)=\dfrac{2x(8-8x)-x^2(-8)}{(8-8x)^2}=\dfrac{16x-16x^2+8x^2}{(8-8x)^2}=\dfrac{16x-8x^2}{(8-8x)^2}[/tex]
