Задача 62109 Нужно найти производную второго порядка...
Условие
Решение
[m]x`_{t}=(\sqrt{1-t^2})`_{t}=\frac{1}{2\sqrt{1-t^2}}\cdot (1-t^2)`_{t}=\frac{1}{2\sqrt{1-t^2}}\cdot (-2t)=\frac{-t}{\sqrt{1-t^2}}[/m]
[m]y`_{t}=(\frac{1}{t})`_{t}=(t^{-1})`_{t}=-1\cdot t^{-2}=-\frac{1}{t^2}[/m]
[m]y``_{xx}=\frac{y``_{tt}\cdot x`_{t}-y`_{t}\cdot x``_{tt}}{x`_{t}}[/m]
[m]x``_{tt}=(x`_{t})`_{t}=\frac{-t}{\sqrt{1-t^2}})`_{t}=\frac{(-t)`_{t}\cdot \sqrt{1-t^2}-(-t)\cdot (\sqrt{1-t^2})`_{t}}{(\sqrt{1-t^2})^2}=
=\frac{-\sqrt{1-t^2}+t\cdot \frac{(-t)}{\sqrt{1-t^2}}}{(\sqrt{1-t^2})^2}=
=\frac{-(1-t^2)-t^2}{(\sqrt{1-t^2})^3}=-\frac{1}{(\sqrt{1-t^2})^3}[/m]
[m]y``_{tt}=(y`_{t})`_{t}=(-1\cdot t^{-2})`_{t}=-(-2)\cdot t^{-3}=\frac{2}{t^3}[/m]
[m]y``_{xx}=\frac{\frac{2}{t^3}\cdot(\frac{-t}{\sqrt{1-t^2}})-(-\frac{1}{(\sqrt{1-t^2})^3} )\cdot (-\frac{1}{t^2}) }{(\frac{-t}{\sqrt{1-t^2}})^3}[/m]
Упрощаем и получаем:
[m]y``_{xx}=\frac{\frac{(-2(1-t^2)-1}{t^2(\sqrt{1-t^2})^3}}{(\frac{-t}{\sqrt{1-t^2}})^3}=\frac{3-2t^2}{t^5}[/m]
О т в е т. [m]y``_{xx}=\frac{3-2t^2}{t^5}[/m]