Найдите производную для функции А) y=x^2-5x+1/x; Б) y=x (x^2-5x+1); B) y=x^3-5x^2+1, в знаменателе х Д) у=х^2/1+х

f (x) ' = ((-1 / 5) * x^3 + 45x^2 + 4x - 1) ' = ((-1 / 5) * x^3) ' + (45x^2) ' + (4x) ' - (1) ' =

(-1 / 5) * 3 * x^2 + 45 * 2 * x + 4 - 0 = (-3 / 5) * x^2 + 90x + 4.

y' = ((cos^2 x) * (1 / x)) ' = (cos^2 x) ' * (1 / x) + (cos^2 x) * (1 / x) ' = (cos x) ' * (cos^2 x) ' * (1 / x) + (cos^2 x) * (1 / x) ' = (-2 * (sin x) * (cos x) * (1 / x)) + (cos^2 x) * (-1 / x^2) = (-2 (sin x) (cos x) / x) + (cos^2 x / x^2).

f (x) ' = ((x^2 + 1) ^ - (1 / 2)) ' = (x^2 + 1) ' * ((x^2 + 1) ^ (-1 / 2)) ' = ((x^2) ' + (1) ') * ((x^2 + 1) ^ (-1 / 2)) ' = 2x * (1 / 2) * (x^2 + 1) ^ (-3 / 2) = x / ((x^2 + 1) ^ (3 / 2)) = x / √ ((x^2 + 1) ^3).

y' = (x^2 * ln (tg √2x)) ' = (x^2) ' * ln (tg √2x) + x^2 * (ln (tg √2x)) ' = (x^2) ' * ln (tg √2x) + x^2 * (tg √2x) ' * (ln (tg √2x)) ' = (x^2) ' * ln (tg √2x) + x^2 * (√2x) ' * (tg √2x) ' * (ln (tg √2x)) ' = 2 * x^ (2 - 1) * ln (tg √2x) + x^2 * √2 * (1 / (cos^2 √2x) * (1 / (tg √2x)) = 2xln (tg √2x) + (√2x^2 / (cos^2 √2x) (tg √2x)).