Question

1) f(x)=x^2/2x-1 2) f(x)=x^2+4/2x-1

Answer 1

Воспользуемся формулой производной частного

$1)~ f'(x)=\bigg(\dfrac{x^2}{2x-1}\bigg)'=\dfrac{(x^2)'\cdot(2x-1)-x^2\cdot (2x-1)'}{(2x-1)^2}=\\ \\ \\ =\dfrac{2x(2x-1)-x^2\cdot 2}{(2x-1)^2}=\dfrac{4x^2-2x-2x^2}{(2x-1)^2}=\dfrac{2x^2-2x}{(2x-1)^2}=\dfrac{2x(x-1)}{(2x-1)^2}$

Аналогично

$2)~ f'(x)=\bigg(\dfrac{x^2+4}{2x-1}\bigg)'=\dfrac{(x^2+4)'\cdot(2x-1)-(x^2+4)\cdot(2x-1)'}{(2x-1)^2}=\\ \\ \\ =\dfrac{2x(2x-1)-2(x^2+4)}{(2x-1)^2}=\dfrac{4x^2-2x-2x^2-8}{(2x-1)^2}=\dfrac{2x^2-2x-8}{(2x-1)^2}$