question about calculus $f(x)=e^x(x^2+x)$, derive $\frac{d^n\,f(x)}{dx^n}$ $f(x)=e^x(x^2+x)$, derive $\dfrac{d^n\,f(x)}{dx^n}$
may use Leibniz formula but i'm not sure:(
 A: It seems a lit bit hard but if you try to find the pattern from n=1 to n=2 ,
so $(e^x (x^2 + x))\prime$ = $e^x (x^2 + 3x +1)$ this is for n egale to 1 you cannot get the pattern , so for $n = 2\quad , (e^x(x^2 + x))\prime\prime = e^x (x^2 + 5x + 4 )$ , to deduce calculate for $n=3$  :
$(e^x(x^2+x))\prime\prime\prime$ = $e^x (x^2 + 7x + 9 )$ , the patern is $\dfrac{d^{n}}{dx^{n}}f(x) = e^x ( x^2 + (2n+1)x +n^2) ...$ If you want to prove the formula you could do the induction .
Hope my answer helped you!
A: Just by calculating $f',f''$ and $f'''$ by hand, you can see the following pattern:
$$\frac{\mathrm{d}^nf}{\mathrm{d}x^n}(x) = e^x(x^2+(2n+1)x+n^2).$$
Now, we can prove this by induction (the base steps are already done if you did the calculations). Assume $\frac{\mathrm{d}^nf}{\mathrm{d}x}(x) = e^x(x^2+(2n+1)x+n^2)$ for some $n\geq 1$, then $$\frac{\mathrm{d}^{n+1}f}{\mathrm{d}x^{n+1}}(x) = e^x(x^2+(2n+1)x+n^2)+e^x(2x+2n+1)$$
The latter can be simplified as $$e^x(x^2+(2(n+1)+1)x+n^2+2n+1)=e^x(x^2+(2(n+1)+1)x+(n+1)^2).$$ This completes the induction step.
A: You can notice that by derivation we will get $e^xP(x)$ with $P$ polynomial, also the degree of $P$ stays unchanged since $(e^x)'$ does not bring some extra $x$.
So we can set $\, f^{(n)}(x)=e^x(a_nx^2+b_nx+c_n)\, $ and try to find a recurrence relation for these sequences.
By derivation we get $\begin{cases}a_{n+1}=a_n&\quad a_0=1\\b_{n+1}=b_n+2a_n&\quad b_0=1\\c_{n+1}=c_n+b_n&\quad c_0=0\end{cases}$
Therefore $a_n$ is constant and $a_n=a_0=1$.
The sequence $b_n$ is telescopic and $$b_{n}=b_0+2\sum\limits_{k=0}^{n-1} a_k=b_{n}=1+2\sum\limits_{k=0}^{n-1} 1=2n+1$$
The sequence $c_n$ is also telescopic and $$c_n=c_0+\sum\limits_{k=0}^{n-1} b_k=0+2\sum\limits_{k=0}^{n-1} k+\sum\limits_{k=0}^{n-1} 1=2\times\frac{(n-1)n}2+n=n^2$$
A: By Newton-Lebniz formula for $D^n[u(x) v(x)]$, we het
$$f(x)=(x^2+x)e^x \implies D^n[e^x(x^2+x)]= (D^n e^x) (x^2+x)+ {n\choose 1} (D^{n-1} e^x) D(x^2+x)+{n \choose 2} (D^{n-2} e^x) D^2(x^2+x)+0$$
$$f^{n}(x)=e^x(x^2+x)+n e^x (2x+1)+n(n-1)e^x=e^x[x^2+(2n+1)x+n^2].$$
