Функције – извод функције 5

Текст задатака објашњених у видео лекцији:

Израчунати извод функције:

Пр.10)   $y = tgx - \sin x$

Пр.11)   $y = \frac{{4x}}{5} - {e^x} + \frac{{\ln x}}{2} - 3arctgx + {5^x}$

Пр.12)   $y = \left( {2 - 2x} \right)\left( {3{x^2} - 4x} \right)$

Пр.13)   $y = {x^5} \cdot \cos x$

Пр.14)   $y = 2xlnx - \frac{{{x^2}}}{4}$

Пр.15)   $y = x - \sin x\cos x$

Пр.10)   $y = tgx - \sin x$

$y' = {\left( {tgx} \right)^\prime } - {\left( {\sin x} \right)^\prime } = \frac{1}{{{{\cos }^2}x}} - \cos x = \frac{{1 - {{\cos }^3}x}}{{{{\cos }^2}x}}$

Пр.11)   $y = \frac{{4x}}{5} - {e^x} + \frac{{\ln x}}{2} - 3arctgx + {5^x}$

$y' = {\left( {\frac{{4x}}{5}} \right)^\prime } - {\left( {{e^x}} \right)^\prime } + {\left( {\frac{{\ln x}}{2}} \right)^\prime } - {\left( {3arctgx} \right)^\prime } + {\left( {{5^x}} \right)^\prime } = $

$ = \frac{4}{5} \cdot 1 - {e^x} + \frac{1}{2} \cdot \frac{1}{x} - 3 \cdot \frac{1}{{1 + {x^2}}} + {5^x}\ln 5 = \frac{4}{5} - {e^x} + \frac{1}{{2x}} - \frac{3}{{1 + {x^2}}} + {5^x}\ln 5$

Пр.12)   $y = \left( {2 - 2x} \right)\left( {3{x^2} - 4x} \right)$

$y' = {\left( {2 - 2x} \right)^\prime }\left( {3{x^2} - 4x} \right) + \left( {2 - 2x} \right){\left( {3{x^2} - 4x} \right)^\prime } = $
$= \left( {2' - {{\left( {2x} \right)}^\prime }} \right)\left( {3{x^2} - 4x} \right) + \left( {2 - 2x} \right)\left( {{{\left( {3{x^2}} \right)}^\prime } - {{\left( {4x} \right)}^\prime }} \right) = $
$= - 2 \cdot \left( {3{x^2} - 4x} \right) + \left( {2 - 2x} \right)\left( {6x - 4} \right) = $
$= - 6{x^2} + 8x + 12x - 8 - 12{x^2} + 8x =  - 18{x^2} + 28x - 8$

Пр.13)   $y = {x^5} \cdot \cos x$

$y' = {\left( {{x^5}} \right)^\prime } \cdot \cos x + {x^5} \cdot {\left( {\cos x} \right)^\prime } = 5{x^4}\cos x - {x^5}\sin x$

Пр.14)   $y = 2xlnx - \frac{{{x^2}}}{4}$

$y' = {\left( {2xlnx} \right)^\prime } - {\left( {\frac{{{x^2}}}{4}} \right)^\prime } = {\left( {2x} \right)^\prime } \cdot lnx + 2x{\left( {lnx} \right)^\prime } - \frac{1}{4} \cdot 2x = $

$ = 2\ln x + 2x \cdot \frac{1}{x} - \frac{1}{2}x = 2\ln x + 2 - \frac{1}{2}x$

Пр.15)   $y = x - \sin x\cos x$

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

$ = {\sin ^2}x + {\sin ^2}x = 2{\sin ^2}x$