Question 1190935
dy/dx=((2x-1)(x^2)'-x^2(2x-1)')/(2x-1)^2


{{{dy/dx=(2x(2x-1)-2x^2)/(2x-1)^2}}}


{{{dy/dx=(2x^2-2x)/(2x-1)^2}}}


{{{dy/dx=0=2x^2-2x}}}   {{{only}}}{{{need}}}{{{numerator}}}


{{{system(for,x=0, and, cross(x=1))}}}

<b>point  (0,0)</b>