Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 815

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$$1)f(x)=\frac{x^2-1}{x^2+1}$$

$$f^{{}^{\prime }}(x)=$$

$$=\frac{{(x^2-1)}^{{}^{\prime }}\bullet (x^2+1)-(x^2-1)\bullet {(x^2+1)}^{{}^{\prime }}}{{(x^2+1)}^2}=$$

$$=\frac{2x\bullet (x^2+1)-(x^2-1)\bullet 2x}{{(x^2+1)}^2}=$$

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

$$=\frac{4x}{{(x^2+1)}^2}$$

$$f^{{}^{\prime }}(1)=\frac{4}{{(1^2+1)}^2}=\frac{4}{2^2}=\frac{4}{4}=1.$$

$$Ответ:1.$$

$$2)f(x)=\frac{2x^2}{1-7x}$$

$$f^{{}^{\prime }}(x)=$$

$$=\frac{{\left(2x^2\right)}^{{}^{\prime }}\bullet (1-7x)-2x^2\bullet (1-7x{)}^{{}^{\prime }}}{(1-7x{)}^2}=$$

$$=\frac{2\bullet 2x\bullet (1-7x)-2x^2\bullet (-7)}{(1-7x{)}^2}=$$

$$=\frac{4x-28x^2+14x^2}{(1-7x{)}^2}=$$

$$=\frac{4x-14x^2}{(1-7x{)}^2}$$

$$f^{{}^{\prime }}(1)=\frac{4-14\bullet 1^2}{(1-7{)}^2}=\frac{4-14}{(-6{)}^2}=$$

$$=-\frac{10}{36}=-\frac{5}{18}.$$

$$Ответ:-\frac{5}{18}.$$