Question

найдите первую и вторую производные
$\sqrt[3]{(\frac{x+1}{x-1})^2 }$

Answer 1

Надеюсь, нигде в арифметике не ошибся. Если что, напиши.

Answer 2

Ответ:

$-\frac{4}{3(x-1)^{2}} \sqrt[3]{\frac{x-1}{x+1}} \quad ;$

$\frac{8}{3(x-1)^{3}} \sqrt[3]{\frac{x-1}{x+1}}-\frac{8}{9(x^{2}-1)^{2}} \sqrt[3]{(\frac{x+1}{x-1})^{2}} \quad ;$

Объяснение:

$\sqrt[3]{(\frac{x+1}{x-1})^{2}}=(\frac{x+1}{x-1})^{\frac{2}{3}}=(\frac{x-1+2}{x-1})^{\frac{2}{3}}=(1+\frac{2}{x-1})^{\frac{2}{3}};$

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

$=\frac{2}{3}(\frac{x+1}{x-1})^{-\frac{1}{3}} \cdot (0+\frac{2' \cdot (x-1)-2 \cdot (x-1)'}{(x-1)^{2}})=\frac{2}{3}(\frac{x-1}{x+1})^{\frac{1}{3}} \cdot (\frac{-2}{(x-1)^{2}})=-\frac{4}{3(x-1)^{2}} \sqrt[3]{\frac{x-1}{x+1}};$

$((1+\frac{2}{x-1})^{\frac{2}{3}})''=(-\frac{4}{3(x-1)^{2}} \sqrt[3]{\frac{x-1}{x+1}})'=-\frac{4}{3} \cdot (\frac{1}{(x-1)^{2}} \cdot (\frac{x-1}{x+1})^{\frac{1}{3}})'=-\frac{4}{3} \cdot ((x-1)^{-2} \cdot$

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

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

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

$+\frac{1}{3(x-1)^{2}} \cdot (\frac{x+1}{x-1})^{\frac{2}{3}} \cdot \frac{2}{(x+1)^{2}})=\frac{8}{3(x-1)^{3}} \sqrt[3]{\frac{x-1}{x+1}}-\frac{8}{9(x^{2}-1)^{2}} \sqrt[3]{(\frac{x+1}{x-1})^{2}};$