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)