Question

Cu ajutorul formulei lui Leibniz sa se calculeze derivatele de ordinul n pentru urmatoarele functii f:D->R:
f(x)=(2x^2+3x+3)e^x

Answer 1

$\displaystyle f^{(n)}(x) = \sum\limits_{i=0}^{n}C_{n}^i(2x^2+3x+3)^{(i)} (e^x)^{(n-i)} =\\ \\ \\= C_{n}^0(2x^2+3x+3){(e^x)}^{(n-0)}+C_{n}^1(2x^2+3x+3)'{(e^x)}^{(n-1)}+\\ \\ +C_{n}^2(2x^2+3x+3)''{(e^x)}^{(n-2)}+C_{n}^3(2x^2+3x+3)'''{(e^x)}^{(n-3)}+0 \\ \\ \\=(2x^2+3x+3)e^x+n\cdot (4x+3)e^x+\dfrac{n(n-1)}{2}\cdot 4\cdot e^x+0+0\\ \\ = e^x(2x^2+3x+3+4nx+3n+2n^2-2n)\\ \\ = e^x\Big[2x^2+(3+4n)x+2n^2+n+3\Big]$