Question

Cho hàm số: y = (x+2)²⁰¹⁹. Tính y⁽⁹⁾(0)

Answer 1

\(\left[\left(x+2\right)^n\right]'=n\left(x+2\right)^{n-1}=\dfrac{n!}{\left(n-1\right)!}.\left(x+2\right)^{n-1}\)

\(\left[\left(x+2\right)^n\right]''=\left(n-1\right)n\left(x+2\right)^{n-2}=\dfrac{n!}{\left(n-2\right)!}\left(x+2\right)^{n-2}\)

Từ đó ta dễ dàng quy nạp được:

\(\left[\left(x+2\right)^n\right]^{\left(k\right)}=\dfrac{n!}{\left(n-k\right)!}\left(x+2\right)^{n-k}\)

Áp vào bài: \(y^{\left(9\right)}\left(0\right)=\dfrac{2019!}{\left(2019-9\right)!}.\left(0+2\right)^{2019-9}=\dfrac{2019!}{2010!}.2^{2010}\)