Задача 69894 Найти dy/dx и d2y/dx2 для заданных...
Условие
a) y=x/(x^2-1)
б) x=cos(t/2) ; y=t-sin(t)
Решение
[m]\frac{dy}{dx} = \frac{1(x^2-1) - x \cdot 2x}{(x^2-1)^2} = \frac{-x^2-1}{(x^2-1)^2}[/m]
[m]\frac{d^2y}{dx^2} = \frac{-2x(x^2-1)^2 + (x^2+1) \cdot 2(x^2-1) \cdot 2x}{(x^2-1)^4} = \frac{-2x(x^2-1) + 4x(x^2+1)}{(x^2-1)^3} =\frac{-2x^3+2x + 4x^3+4x}{(x^2-1)^3} =\frac{2x^3+6x}{(x^2-1)^3}[/m]
б) [m]x= cos \frac{t}{2};\ y = t-sin(t)[/m]
[m]x'_t=\frac{dx}{dt} = \frac{1}{2}(-sin \frac{t}{2}) = -\frac{1}{2} sin \frac{t}{2} = -1/2 \cdot sin(t/2)[/m]
[m]y'_t=\frac{dy}{dt} = 1-cos(t)[/m]
[m]y'_x=\frac{dy}{dx} = \frac{dy}{dt} : \frac{dx}{dt} = -\frac{1-cos(t)}{1/2 \cdot sin(t/2)}=-2 \cdot \frac{1-cos(t)}{sin(t/2)}[/m]
[m](y'_x)'_t = -2\frac{-(-sin(t))sin(t/2) - (1-cos(t))cos(t/2)(1/2)}{sin^2(t/2)} = -2\frac{sin(t)sin(t/2) - 0,5cos(t/2) + 0,5cos(t)cos(t/2)}{sin^2(t/2)} [/m]
[m]\frac{(y'_x)'_t}{x'_t} = -2\frac{sin(t)sin(t/2) - 0,5cos(t/2) + 0,5cos(t)cos(t/2)}{sin^2(t/2)} : (-1/2 \cdot sin(t/2)) =[/m]
[m]= 4\frac{sin(t)sin(t/2) - 0,5cos(t/2) + 0,5cos(t)cos(t/2)}{sin^3(t/2)}[/m]