Вычислите dy/dx и d^2y/dx^2, если функция y (x) задана параметрически, x=(1+cos^2t) ^2y=cost/sin^2t

X=(1+(cos (t) ^2) ^2y=cos (t) / (sin (t) ^2Решение. Найдем вначале первую производнуюdy/dx=(dy/dt) / (dx/dt) Отдельно находим производные xt' и yt'dx/dt=2 (1+(cos (t) ^2)*2cos (t)*(-sin ()=-4 (1+(cos (t) ^2)*cos (t)*sin (t) dy/dt=(- (sin (t) ^3-2 (cos (t) ^2*sin (t) / (sin (t) ^4=- (sin (t) ^2+2 (cos (t) ^2) / (sin (t) ^3=- (1+(cos (t) ^2) / (sin (t) ^3 Следовательно: dy/dx=[- (1+(cos (t) ^2) / (sin (t) ^3]/[-4 (1+(cos (t) ^2)*cos (t)*sin (t) ]=1/ (4*(sin (t) ^4*cos (t) Найдем yx' (вторую производную): y’’=[d (dy/dx) /dt]/[dx/dt] d (dy/dx) /dt=(1/4)*(sin (t) ^ (-4)*(cos (t) ^ (-1) ’=(1/4)*(-4)*(sin (t) ^ (-5)*cos (t)*(cos (t) ^ (-1)+(sin (t) ^ (-4)*(-1) (cos (t) ^ (-2)*sin (t)=(1/4)*(-4/ (sin (t) ^ (5) – 1/[ (sin (t) ^ (3)*(cos (t) ^ (2) ])=(-1/4)*(4 (cos (t) ^2+(sin (t) ^2) / (sin (t) ^5*(cos (t) ^2)=- (3 (cos (t) ^2+1) / (4 (sin (t) ^5*(cos (t) ^2) Тогда y’’=- (3 (cos (t) ^2+1) / (4 (sin (t) ^5*(cos (t) ^2) / (-4 (1+(cos (t) ^2)*cos (t)*sin (t)=(3 (cos (t) ^2+1) / (16*(sin (t) ^6*(cos (t) ^3*(1+(cos (t) ^2)