Найти производные функции в точке x0, если: 1) y=lntg3*x; x0=π/4 2) y=cos^2 (5*x); x0=π/2 3) y=e^ (2*x^2+3); x0=1 можно без решения а только ответы

f (x) ' = ((2ln x + 3) ^ (-1)) ' = (2ln x + 3) ' * ((2ln x + 3) ^ (-1)) ' = ((2ln x) ' + (3) ') * ((2ln x + 3) ^ (-1)) ' = (2 * (1 / x) + 0) * (-1) * (2ln x + 3) ^ (-1 - 1) = (2 / x) * (-1) * (2ln x + 3) ^ (-2) = (-2) / (x * (2ln x + 3) ^2).

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

y' (3π / 4) = 3 * cos (3 * (3π / 4)) + 3 * sin (3 * (3π / 4)) = 3 * cos (9π / 4) + 3 * sin (9π / 4) = 3 * cos (2π + (π / 4)) + 3 * sin (2π + (π / 4)) = 3 * cos (π / 4) + 3 * sin (π / 4) = 3 * (cos (π / 4) + sin (π / 4)) = 3 * ((√2 / 2) + (√2 / 2)) = 3 * √2 = 3√2.

y' = (cos (arcsin x)) ' = (arcsin x) ' * (cos (arcsin x)) ' = (1 / √ (1 - х^2)) * (-sin (arcsin x)) = (-sin (arcsin x)) / √ (1 - х^2).