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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?

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Try a few derivatives by hand and try to notice the pattern. Note that this is a product of two functions - you will get the general answer from Leibnitz Rule if you try to differentiate a product of $f(x)g(x)$ by hand some number of times and you will see that nth derivative of that has a very nice form. – user116017 Dec 19 '13 at 19:10
As usser116017 suggested, you have to notice the pattern – John Dec 19 '13 at 19:11
I edited my answer showing the use of the Leibniz rule. – user63181 Dec 19 '13 at 19:33
up vote 30 down vote accepted

Do you know the Leibniz formula for derivative $$(fg)^{(n)}=\sum_{k=0}^n{n\choose k}f^{(k)}g^{(n-k)}?$$


Since the third derivative of $x^2$ is $0$ so $$(x^2e^x)^{(10)}=\sum_{k=0}^2{10\choose k}(x^2)^{(k)}(e^x)^{(10-k)}\\={10\choose 0}x^2e^x+{10\choose1}\times 2xe^x+{10\choose 2}\times2e^x=(x^2+20x+90)e^x$$

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I don't, but I think it's the key to my success! Would you care to explain it and how it applies to this question? – Mr Croutini Dec 19 '13 at 19:17
Here it's interesting because with $f(x)=x^2$ and $g(x)=e^x$, you have $f^{(k)}(x)=0$ for $k>2$, and $g^{(n-k)}(x)=e^x$ for all $k$. The formula thus simplifies greatly. You just have to keep three terms ($k=0,1,2$). – Jean-Claude Arbaut Dec 19 '13 at 19:19
I've got it! Thanks! I was overcomplicating it in my head for some reason! Also, your solution makes much more sense than the one I was given! – Mr Croutini Dec 19 '13 at 21:11
You're welcome Mr Croutini. – user63181 Dec 19 '13 at 21:12
That's a fine answer! +1 – amWhy Dec 20 '13 at 13:21

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


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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 $\ a,b,c=1,0,0\ \Rightarrow \dfrac{d^{10}}{{dx}^{10}}\left(x^2 e^x\right)\, =\, (x^2+ 20x+90)\,e^x$

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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:


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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

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 $x=a$.

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+1 for the fun part. – achille hui Dec 19 '13 at 20:34

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.

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Hint: $$(fg)^{(n)}= \sum_{k=0}^n {n \choose k} f^{(n-k)}g^{(k)}$$

One could prove this with Mathematical Induction.

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This is Leibniz' Rule. – robjohn Dec 19 '13 at 19:14

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