Задача 70458 Найти y' и у'' { x = (ln t) / t { y = t...
Условие
{ x = (ln t) / t
{ y = t ln t
Решение
{ y = t*(ln t)
Производная от функции, заданной параметрически.
{ x'_(t) = (1/t*t - 1*ln t)/t^2 = (1 - ln t)/t^2
{ y'_(t) = 1*ln t + t*1/t = ln t + 1
[m] y'_(x) = y'_{t} : x'_{t} = \frac{ln\ t + 1}{(1 - ln\ t)/t^2} = \frac{t^2(1 + ln\ t)}{1 - ln\ t} [/m]
Производная 2 порядка:
y''_(xx) = (y'_(x))'_(t)/x'_(t)
[m] (y'_{x})'_{t} = \frac{2t(1 + ln\ t)(1 - ln\ t) + t^2 \cdot 1/t(1 - ln\ t) - t^2(1 + ln\ t)(-1/t)}{(1-ln\ t)^2} = [/m]
[m] =\frac{(2t+ 2t \cdot ln\ t + t)(1 - ln\ t) + t(1 + ln\ t)}{(1-ln\ t)^2} = \frac{(3t+ 2t \cdot ln\ t)(1 - ln\ t) + t(1 + ln\ t)}{(1-ln\ t)^2} = [/m]
[m] =\frac{3t+ 2t \cdot ln\ t - 3t \cdot ln\ t -2t \cdot ln^2\ t + t+ t \cdot ln\ t}{(1-ln\ t)^2} = \frac{4t - 2t \cdot ln^2\ t}{(1-ln\ t)^2} = \frac{2t \cdot (2 - ln^2\ t)}{(1-ln\ t)^2} [/m]
[m](y'_{x})'_{t} = \frac{2t \cdot (2 - ln^2\ t)}{(1-ln\ t)^2}[/m]
[m]y''_{xx} = \frac{(y'_{x})'_{t}}{x'_{t}} = \frac{2t \cdot (2 - ln^2\ t)}{(1-ln\ t)^2} : \frac{1 - ln\ t}{t^2} = \frac{2t \cdot (2 - ln^2\ t)}{(1-ln\ t)^2} \cdot \frac{t^2}{1 - ln\ t} = \frac{2t^3 \cdot (2 - ln^2\ t)}{(1-ln\ t)^3}[/m]