Question

sa se deriveze fractia : x²-2x+4 supra x²-x

Answer 1

$( \frac{x^2-2x+4}{x^2-x} )'= \frac{(2x-2)(x^2-x)-(2x-1)(x^2-2x+4)}{(x^2-x)^2} = \\ = \frac{2x^3-2x^2-2x^2+2x-2x^3+4x^2-8x+x^2-2x+4}{(x^2-x)^2}= \frac{x^2-8x+4}{(x^2-x)^2}$

Regula de derivare a unei fractii:

$(\frac{f(x)}{g(x)}) '= \frac{f'(x)g(x)-g'(x)f(x)}{g(x)^2}$