Найти f´(3) и f´(1) Если :f(x)=1/x3 + 1/x2

f(x)=1/x3 + 1/x2=x⁻³ + x⁻² f'(x) = -3x⁻⁴ -2x⁻³ = -3/х⁴ - 2/х³ f'(3) = -3/81 - 2/27 = -1/27 -2/27 = -3/27 = -1/9 f'(1) = -3 -2 = -5