$y = (x^2 + 2x - 3)e^x$

Đặt $\begin{cases}h(x) = x^2 + 2x - 3\\g(x)= e^x\end{cases}$

$\Rightarrow y = h(x).g(x)$

Ta có:

$g^{(n)}(x)= (e^x)^{(n)}= e^x$

$h'(x)= 2x - 2$

$h''(x)= 2$

$h'''(x)= 0$

$\Rightarrow h^{(n)}(x)= 0,\ \ (n\geqslant 3)$

Áp dụng công thức $\text{Leibniz}$ ta được:

$y^{(2020)}= \displaystyle\sum\limits_{k=0}^{2020}C_{2020}^kh^{(k)}(x).g^{(2020 - k)}(x)$

$\qquad = C_{2020}^0h(x).g^{(2020)}(x) + C_{2020}^1h'(x).g^{(2019)}(x) + C_{2020}^2h''(x).g^{(2018)}(x)$

$\qquad = (x^2 + 2x -3)e^x + 2020(2x +2)e^x+ 2039190.2e^x$

$\qquad = (x^2 + 4042x + 4082417)e^x$

y=(x2+2x−3)exy=(x2+2x−3)ex

Đặt {h(x)=x2+2x−3g(x)=ex{h(x)=x2+2x−3g(x)=ex

⇒y=h(x).g(x)⇒y=h(x).g(x)

Ta có:

g(n)(x)=(ex)(n)=exg(n)(x)=(ex)(n)=ex

h′(x)=2x−2h′(x)=2x−2

h′′(x)=2h″(x)=2

h′′′(x)=0h‴(x)=0

⇒h(n)(x)=0,  (n⩾3)⇒h(n)(x)=0,  (n⩾3)

Áp dụng công thức LeibnizLeibniz ta được:

y(2020)=2020∑k=0Ck2020h(k)(x).g(2020−k)(x)y(2020)=∑k=02020C2020kh(k)(x).g(2020−k)(x)

=C02020h(x).g(2020)(x)+C12020h′(x).g(2019)(x)+C22020h′′(x).g(2018)(x)=C20200h(x).g(2020)(x)+C20201h′(x).g(2019)(x)+C20202h″(x).g(2018)(x)

=(x2+2x−3)ex+2020(2x+2)ex+2039190.2ex=(x2+2x−3)ex+2020(2x+2)ex+2039190.2ex

=(x2+4042x+4082417)ex