Question

Найдите производную g(x)= x^2-4x/x+1; x0= -2​

Answer 1

$g(x)=\dfrac{x^2-4x}{x+1}\ \ ,\ \ \ x_0=-2\\\\\\g'(x)=\dfrac{(2x-4)(x+1)-(x^2-4x)\cdot 1}{(x+1)^2}=\dfrac{2x^2-2x-4-x^2+4x}{(x+1)^2}=\\\\\\=\dfrac{x^2+2x-4}{(x+1)^2}\\\\\\g'(-2)=\dfrac{4-4-4}{(-1)^2}=4$