Функције – извод функције 6

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Пр.16)   $y = \frac{{{x^2}}}{{{x^2} + 1}}$

Пр.17)   $y = \frac{{{e^x}}}{{{x^4}}}$

Пр.18)   $y = \frac{{\ln x + 1}}{{\ln x}}$

Пр.19)   $y = \frac{{1 - \cos x}}{{1 + \cos x}}$

Пр.20)   $y = \frac{{arctgx}}{x} - \frac{{\ln x}}{2}$

Пр.16)   $y = \frac{{{x^2}}}{{{x^2} + 1}}$

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

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

Пр.17)   $y = \frac{{{e^x}}}{{{x^4}}}$

$y' = \frac{{{{\left( {{e^x}} \right)}^\prime }{x^4} - {e^x}{{\left( {{x^4}} \right)}^\prime }}}{{{x^8}}} = \frac{{{e^x} \cdot {x^4} - {e^x} \cdot 4{x^3}}}{{{x^8}}} = \frac{{{e^x} \cdot {x^3}\left( {x - 4} \right)}}{{{x^8}}} = \frac{{{e^x} \cdot \left( {x - 4} \right)}}{{{x^5}}}$

Пр.18)   $y = \frac{{\ln x + 1}}{{\ln x}}$

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

$ = \frac{{ - \frac{1}{x}}}{{{{\ln }^2}x}} = \frac{1}{{x{{\ln }^2}x}}$

Пр.19)   

$y = \frac{{1 - \cos x}}{{1 + \cos x}}$

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

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

Пр.20)   $y = \frac{{arctgx}}{x} - \frac{{\ln x}}{2}$

$y' = {\left( {\frac{{arctgx}}{x}} \right)^\prime } - {\left( {\frac{{\ln x}}{2}} \right)^\prime } = \frac{{{{\left( {arctgx} \right)}^\prime }x - x'arctgx}}{{{x^2}}} - \frac{1}{2} \cdot \frac{1}{x} = $

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

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

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