Question

use Leibnitz theorem to evaluate the fourth derivative of \\(\\left(2x^{3}+x^{2}+x+2\\right)e^{2x}\\)

Answer 1

Lebnicz formula for the product

$(f(x)g(x)) ^{(n)}= \sum\limits _{r=0}^{n}C^r_n(f )^{(n-r)}(x) (g )^{(r)}(x)$

$f(x)= 2x^{3}+x^{2}+x+2, g(x)=e^{2x}$

$(f(x)g(x)) ^{(4)}=((2x^{3}+x^{2}+x+2)e^{2x}) ^{(4)} = \\ = \sum \limits _{r=0}^{4} C^r_4(2x^{3}+x^{2}+x+2) ^{(4-r)} (e^{2x} )^{(r)}$ =

$=\sum\limits _{r=0}^{n}C^r_4((2x^{3}+x^{2}+x+2) )^{(4-r)}(x) (e^{2x} )^{(r)}=\\=C^0_4((2x^{3}+x^{2}+x+2) )^{(4)}(x) (e^{2x} )+\\ +C^1_4((2x^{3}+x^{2}+x+2) )^{(3)}(x) (e^{2x} )^{(1)}+\\+C^2_4((2x^{3}+x^{2}+x+2) )^{(2)}(x) (e^{2x} )^{(2)}+\\+C^3_4((2x^{3}+x^{2}+x+2) )^{(1)}(x) (e^{2x} )^{(3)}+\\+C^4_4((2x^{3}+x^{2}+x+2) )(x) (e^{2x} )^{(4)}$

$\frac{}{}$   $C^r_n=\frac{n!}{(n-k)! k!} ​$

$C^0_4=\frac{4!}{(4)! 0!}=1 \\ C^1_4=\frac{4!}{(3)! 1!}=4\\ C^2_4=\frac{4!}{(2)! 2!}=6\\C^3_4=\frac{4!}{(1)! 3!}=4\\C^4_4=\frac{4!}{(0)! 4!}=1$

$(2x^{3}+x^{2}+x+2) ^{(1)}=6x^{2}+2x^+1\\ (2x^{3}+x^{2}+x+2) ^{(2)}=(6x^{2}+2x+1)^{(1)}=12x+2\\ (2x^{3}+x^{2}+x+2) ^{(3)}=(12x+2)^{(1)}=12\\ (2x^{3}+x^{2}+x+2) ^{(4)}=0$

$(e^{2x} )^{(1)}=2e^{2x} \\ (e^{2x} )^{(2)}=4e^{2x}\\ (e^{2x} )^{(3)}=8e^{2x}\\ (e^{2x} )^{(4)}=16e^{2x}$

So

$((2x^{3}+x^{2}+x+2)e^{2x}) ^{(4)}=\\=1*0*e^{2x}+4*2*12 e^{2x}+ \\+6*4*(12x+2) e^{2x}+ \\+4*8*(6x^{2}+2x+1) e^{2x}+\\+1*16 (2x^{3}+x^{2}+x+2) e^{2x}=\\=e^{2x}(32x^3+208x^2+368x+208)$

Answer:

$\frac{d^4}{dx^4} ((2x^{3}+x^{2}+x+2)e^{2x})=e^{2x}(32x^3+208x^2+368x+208).$