`y = x.ln (1 + 2x)`
Áp dụng Leibneiz, ta có:
`y`
`= sum_{k = 0}^{10} C_{10}^{k}.x^{(k)}.[ln (1 + 2x)]^{(10 - k)}`
`= C_{10}^{0}.x.[ln (1 + 2x)]^{(10)} + C_{10}^{1}.1.[ln (1 + 2x)]^{(9)}`
Có: `[ln (1 + 2x)]' = 2/(1 + 2x)`
`-> [ln (1 + 2x)]^{(9)} = 2.((1)/(1 + 2x))^{(8)} = 2.(2^{8}.(-1)^{8}.8!)/((1 + 2x)^{9})`
`-> y^{(10)} (0) = C_{10}^{1}.(2^{9}.8!)/((1 + 2.0)^{9}) = C_{10}^{1}.2^{9}.8!`
