Question

пожалуйста помогите 1 и 3

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Answer 1

Объяснение:

$1)\\y=\frac{x-1}{x+1} \ \ \ \ \ x_0=2\\\\y'(x)=(\frac{x-1}{x+1})'=\frac{(x-1)'*(x+1)-(x-1)*(x+1)'}{(x+1)^2} =\frac{x+1-(x-1)}{(x+1)^2}=\frac{x+1-x+1}{(x+1)^2}=\frac{2}{(x+1)^2}.\\y'(2)=\frac{2}{(2+1)^2}=\frac{2}{3^2}=\frac{2}{9}.\\3)\ \\ y=\frac{x+2}{x-4}\ \ \ \ x_0=5\\y'(x)=\frac{(x+2)'*(x-4)-(x+2)*(x-4)'}{(x-4)^2}=\frac{x-4-(x-2)}{(x-4)^2}=\frac{x-4-x-2}{(x-4)^2}=-\frac{6}{(x-4)^2}.\\y'(5)=-\frac{6}{(5-4)^2}=-\frac{6}{1^2}=-\frac{6}{1}=-6.$