Ответ: d. 1.25
Объяснение:
[tex]y=\frac{x^2+2x}{x+1} \\ u=x^2+2x \\v=x+1\\y' =\frac{u'v-v'u}{v^2}\\ u' = 2x+2\\v'=1\\y=\frac{(2x+2)(x+1)-(x^2+2x)}{(x+1)^2} =\frac{2x^2+4x+2-x^2-2x}{(x+1)^2}= \frac{x^2+2x+2}{(x+1)^2}\\[/tex]
[tex]y'(1)=\frac{1+2+2}{(1+1)^2}=\frac{5}{4} =1.25[/tex]
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Ответ: d. 1.25
Объяснение:
[tex]y=\frac{x^2+2x}{x+1} \\ u=x^2+2x \\v=x+1\\y' =\frac{u'v-v'u}{v^2}\\ u' = 2x+2\\v'=1\\y=\frac{(2x+2)(x+1)-(x^2+2x)}{(x+1)^2} =\frac{2x^2+4x+2-x^2-2x}{(x+1)^2}= \frac{x^2+2x+2}{(x+1)^2}\\[/tex]
[tex]y'(1)=\frac{1+2+2}{(1+1)^2}=\frac{5}{4} =1.25[/tex]