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Old 16.10.2004, 08:19   #16
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как вам угодно, коллега.
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Old 16.10.2004, 08:26   #17
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f(x) = (x+5)^3 * e^(x/2)^3

чему равно d(f(x))/dx ?
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Old 16.10.2004, 08:36   #18
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d(f(x))/dx=f'(x)=lim(deltaX->0)[(f(x+deltaX)-f(x))/deltaX].
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Old 16.10.2004, 14:44   #19
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это наверное равно 19
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Old 22.09.2017, 03:23   #20
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Quote:
Originally Posted by RAMZES View Post
f(x) = (x+5)^3 * e^(x/2)^3

чему равно d(f(x))/dx ?
Замечательная весч Вольфрам Математика.

d(f(x))/dx=3 у^(3 x/2) (5 + x)^2 + 3/2 у^(3 x/2) (5 + x)^3

А вот график:


d(f(x))/dx=3 e^(3 x/2) (5 + x)^2 + 3/2 e^(3 x/2) (5 + x)^3
гы, лол.
вместо у экспонента.

Last edited by AvDav; 22.09.2017 at 03:23.
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Old 06.10.2017, 14:07   #21
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На самом деле легко выводится если применить правило вычисления производных двух функций.
(U*V)' = U'V + V'U
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Old 03.11.2017, 19:03   #22
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Надо же, как оперативно удаляют картинки.
Ителекчуал проперти, хуле.
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Old 04.11.2017, 18:49   #23
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Quote:
Originally Posted by AvDav View Post
Надо же, как оперативно удаляют картинки.
Ителекчуал проперти, хуле.
Ситуация меняется с каждым днём.
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Old 19.07.2018, 14:54   #24
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А вот и решение при помощи пакета MATLAB.
d(f)/dx = 3*exp((3*x)/2)*(x + 5)^2 + (3*exp((3*x)/2)*(x + 5)^3)/2
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Old 25.10.2018, 15:42   #25
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Вольфрам Математика с Матлабом согласна
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Old 05.12.2018, 14:38   #26
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Quote:
Originally Posted by Sauron View Post
потрясающе
жаль нобелевскую премию по математике на дают
нобелевскую не дают, зато дают абелевскую.
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