Найти f' (x)

a) f (x) = (x-3) ^4;

б) f (x) = (2x+1) ^2;

в) y=2√x * (3x+4) ;

г) y=x*ctgx;

д) y=2-x^2/x^2+2;

е) y=3-√x/x^2-4.

А) f' (x) = 4 (x - 3) ³

б) f' (x) = 4 (2x + 1)

в) y (x) = 6x^ (3/2) + 8x^ (1/2)

y' (x) = 9x^ (1/2) + 4x^ (-1/2) = 9√x + 4/√x

г) y' (x) = ctgx + x * (-1/sin²x) = - x*ctgx/sin²x = - xcosx/sin³x

д) y' (x) = (-2x (x² + 2) - 2x (2 - x²)) / (x² + 2) ² = (-2x³ - 4x - 4x + 2x³) / (x² + 2) ² = - 8x / (x² + 2) ²

е) y' (x) = (-1/2 * x^ (-1/2) (x² - 4) - 2x (3 - √x)) / (x² - 4) ² = (-x² + 4 - 4x^ (3/2) (3 - √x)) / (2√x (x² - 4) ²) = (-x² + 4 - 12x^ (3/2) + 4x²)) / (2√x (x² - 4) ²) = (4 + 3x² - 12x^ (3/2))) / (2√x (x² - 4) ²)