Функције – извод функције 8

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Пр.7)   $y = \sqrt {\frac{{x + 1}}{{x - 1}}} $

Пр.8)   $y = \ln \sqrt[3]{{2x - 1}}$

Пр.9)   $y = {\left( {x + {{\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)}^2}} \right)^2}$

Пр.7)  $y = \sqrt {\frac{{x + 1}}{{x - 1}}} $

$y' = {\left( {\sqrt {\frac{{x + 1}}{{x - 1}}} } \right)^\prime } = {\left( {{{\left( {\frac{{x + 1}}{{x - 1}}} \right)}^{\frac{1}{2}}}} \right)^\prime } = \frac{1}{2}{\left( {\frac{{x + 1}}{{x - 1}}} \right)^{\frac{1}{2} - 1}}{\left( {\frac{{x + 1}}{{x - 1}}} \right)^\prime } = $

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

$ = \frac{1}{2}{\left( {\frac{{x + 1}}{{x - 1}}} \right)^{ - \frac{1}{2}}} \cdot \frac{{\left( {x - 1} \right) - \left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}} = \frac{1}{2}{\left( {\frac{{x + 1}}{{x - 1}}} \right)^{ - \frac{1}{2}}} \cdot \frac{2}{{{{\left( {x - 1} \right)}^2}}} = \sqrt {\frac{{x + 1}}{{x - 1}}}  \cdot \frac{1}{{{{\left( {x - 1} \right)}^2}}}$

Пр.8)   $y = \ln \sqrt[3]{{2x - 1}}$

$y' = \frac{1}{{\sqrt[3]{{2x - 1}}}} \cdot {\left( {{{\left( {2x - 1} \right)}^{\frac{1}{3}}}} \right)^\prime } = \frac{1}{{\sqrt[3]{{2x - 1}}}} \cdot \frac{1}{3} \cdot {\left( {2x - 1} \right)^{\frac{1}{3} - 1}} \cdot {\left( {2x - 1} \right)^\prime } = $

$ = \frac{1}{{\sqrt[3]{{2x - 1}}}} \cdot \frac{1}{3} \cdot {\left( {2x - 1} \right)^{ - \frac{2}{3}}} \cdot 2 = \frac{2}{3} \cdot \frac{1}{{{{\left( {2x - 1} \right)}^{\frac{1}{3}}}{{\left( {2x - 1} \right)}^{\frac{2}{3}}}}} = \frac{2}{{3\left( {2x - 1} \right)}}$

Пр.9)   $y = {\left( {x + {{\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)}^2}} \right)^2}$

$y' = {\left( {{{\left( {x + {{\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)}^2}} \right)}^2}} \right)^\prime } = $

$ = 2 \cdot \left( {x + {{\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)}^2}} \right) \cdot {\left( {x + {{\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)}^2}} \right)^\prime } = $

$ = 2 \cdot \left( {x + {{\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)}^2}} \right) \cdot \left( {1 + {{\left( {{{\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)}^2}} \right)}^\prime }} \right) = $

$ = 2 \cdot \left( {x + {{\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)}^2}} \right) \cdot \left( {1 + 2\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right) \cdot {{\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)}^\prime }} \right) = $

$ = 2 \cdot \left( {x + {{\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)}^2}} \right) \cdot \left( {1 + 2\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right) \cdot \left( {1 + {{\left( {{{\left( {{x^2} + 1} \right)}^2}} \right)}^\prime }} \right)} \right) = $

$ = 2 \cdot \left( {x + {{\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)}^2}} \right) \cdot \left( {1 + 2\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)\left( {1 + 2\left( {{x^2} + 1} \right){{\left( {{x^2} + 1} \right)}^\prime }} \right)} \right) = $

$ = 2 \cdot \left( {x + {{\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)}^2}} \right) \cdot \left( {1 + 2\left( {x + {{\left( {{x^2} + 1} \right)}^2}} \right)\left( {1 + 2\left( {{x^2} + 1} \right)2x} \right)} \right)$