Verified answer

Объяснение:

[tex]y=\frac{(x-3)^2}{4*(x-1)}\\y'=( \frac{(x-3)^2}{4*(x-1)})'=\frac{((x-3)^2)'*4*(x-1)-(x-3)^2*4*(x-1)'}{(4*(x-1))^2}= \frac{2*4*(x-3)*(x-1)-4*(x-3)^2}{16*(x-1)^2}=\\=\frac{4*(2x^2-8x+6-x^2+6x-9)}{16*(x-1)^2} =\frac{x^2-2x-3}{4*(x-1)^2} .[/tex]