Find the tenth derivative of $x^2e^x$ If $$f(x) = x^2e^x$$
find $$f^{({10})}x$$
I'm not familiar with doing the series expansion for this and the solution that was provided to me did not help me at all, as it wasn't explained. I think it's "Higher Leibnitz Rule", but I'm unsure.
I could differentiate the function 10 times, but the solutions uses summation. How would I go about doing this?
 A: Hint $\ \ f = (ax^2\!+bx+c)\,e^x\Rightarrow\, f' = (ax^2\!+(b+2a)x+c+b)\,e^x.\, $ Abbreviating $f\mapsto f'$ as
$$(a,b,c)\ \mapsto\ (a, b+2a,c+b)$$
We iterate it $n$ times to get $f^{(n)}.$ Looking for patterns in the first few values we have
$$\begin{eqnarray}
f^{(1)} = (a,&&b+2a,&&c\, +\, b)\\
f^{(2)} = (a,&&b+4a,&&c+2b+2a)\\
f^{(3)} =(a,&&b+6a,&&c+3b+6a)\\
f^{(4)} = (a,&&b+8a,&&c+4b+12a)\\
\end{eqnarray}$$
Hence the pattern appears to be  $\ f^{(n)} = (a,\,b\!+\!2na,\,c+\!nb+n(n\!-\!1)a) $ 
This is immediately proved by induction, the induction step being
$$  \begin{eqnarray} &&(a,&&b+2na,&&c+nb+n(n\!-\!1)a) &[\,= f^{(n)}\,]&\\
    + &&(0,&&\quad\quad\ 2a,&&\quad\quad  b+2na)\\
   = &&(a,&&b+2(n\!+\!1)a,&&c+(n\!+\!1)b+(n\!+\!1)na)\ \ \ &[\,= f^{(n+1)}\,]&\end{eqnarray}$$
Therefore $\ \ \dfrac{d^n}{{dx}^n}\left((ax^2\!+bx\!+\!c)e^x\right)\, =\, (ax^2\!+(b\!+\!2na)x+c\!+\!nb\!+\!n(n\!-\!1)a)\,e^x$
so $\ a,b,c=1,0,0\ \Rightarrow  \dfrac{d^{10}}{{dx}^{10}}\left(x^2 e^x\right)\, =\, (x^2+ 20x+90)\,e^x$
A: First derivative: $$x^2e^x+2xe^x$$
Second derivative: $$x^2e^x+4xe^x+2e^x$$
Third Derivaitve: $$x^2e^x+6xe^x+6e^x$$
I think you can see the pattern now:
$$f^{(n)}=x^2e^x+(2n)e^x+(n^2-n)e^x$$
A: For fun, since good ways have been described, we do not so good. First a small change of notation. Let $f(x)=x^2e^x$. We find $f^{(10)}(a)$, or more generally
$f^{(n)}(a)$.
Expand $e^x$ in a power series about $x=a$. We get 
$$e^x=\sum_0^\infty \frac{e^a}{n!}(x-a)^n.\tag{1}$$ 
Expand $x^2$ about $x=a$. We get 
$$x^2=a^2+2a(x-a)+(x-a)^2.\tag{2}$$
For $n\ge 2$, the coefficient of $(x-a)^n$ in the product of (1) and (2) is
$$e^a\left(\frac{1}{n!}a^2+\frac{2}{(n-1)!}a+\frac{1}{(n-2)!}\right).$$
The above coefficient is $\frac{1}{n!}$ times the derivative of $x^2f(x)$ at $x=a$. It follows that the required derivative is
$$e^a\left(a^2 +2na+n(n-1)\right).$$
Remark: By expanding $x^k$ about $x=a$, we can in the same way find the $n$-th derivative of $x^ke^x$ at $x=a$. 
A: One trick to compute derivatives of function containing an exponential factor $e^{\alpha x}$ is the formal identity
$$e^{-\alpha x} \frac{d}{dx} e^{\alpha x} = \frac{d}{dx} + \alpha$$
What this means is if you put any differentiable function $f(x)$ on the RHS on both sides of above formal identity, you get back a valid identity:
$$ e^{-\alpha x}\frac{d}{dx}\left[e^{\alpha x} f(x)\right] = (\frac{d}{dx} + \alpha) f(x) = f'(x) + \alpha f(x)$$
Notice on the space of continuous functions, the actions of multiplying a factor $e^{\alpha x}$ and $e^{-\alpha x}$ are inverse operations to each other. This leads us to another formal identity
$$e^{-\alpha x} \frac{d^n}{dx^n} e^{\alpha x} = \left( e^{-\alpha x}\frac{d}{dx} e^{\alpha x}\right)^n = \left(\frac{d}{dx} + \alpha\right)^n = \sum_{k=0}^n \binom{n}{k}\alpha^{n-k} \frac{d^k}{dx^k}$$
If you put $\alpha = 1, n = 10$ and $x^2$ on the RHS of above formal identity, you get
$$e^{-x} \frac{d^{10}}{dx^{10}} \left[e^x x^2\right] = \sum_{k=0}^{10}\binom{10}{k}\frac{d^k}{dx^k} x^2 =
x^2 + \binom{10}{1} (2x) + \binom{10}{2} 2
\\
\implies \quad \frac{d^{10}}{dx^{10}} \left[e^x x^2\right] = e^x \left( x^2 + 20x + 90 \right)
$$
Please note that this is not a rigorous derivation of the result. It is a trick to help you to skip the induction steps and get the correct answer quickly.
A: Do you know the Leibniz formula for derivative
$$(fg)^{(n)}=\sum_{k=0}^n{n\choose k}f^{(k)}g^{(n-k)}?$$
Added:
Since the third derivative of $x^2$ is $0$ so
$$(x^2e^x)^{(10)}=\sum_{k=0}^2{10\choose k}(x^2)^{(k)}(e^x)^{(10-k)}\\={10\choose 0}x^2e^x+{10\choose1}\times 2xe^x+{10\choose 2}\times2e^x=(x^2+20x+90)e^x$$
A: Hint: The derivative of $P(x)e^x$ is $(P'(x)+P(x))e^x$.
So, basically, you have to apply 10 times the transformation $P \to P' + P$, which should not be difficult with the polynomial $P(x)=x^2$:
$$x^2$$
$$x^2+2x$$
$$(x^2+2x)+(2x+2)=x^2+4x+2$$
$$(x^2+4x+2)+(2x+4)=x^2+6x+6$$
$$(x^2+6x+6)+(2x+6)=x^2+8x+12$$
$$(x^2+8x+12)+(2x+8)=x^2+10x+20$$
The last line is the fifth derivative, with factor $e^x$ removed.
Now, some induction will help you prove that
$$(x^2e^x)^{(n)}=(x^2+2nx+n(n-1))e^x$$
A: Hint: $$(fg)^{(n)}= \sum_{k=0}^n {n \choose k} f^{(n-k)}g^{(k)}$$
One could prove this with Mathematical Induction.
