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Find the $n^{th}$ derivative of $$f(x) = e^x\cdot x^m$$

If i am not wrong i have following

$1^{st}$ Derivative: $e^x\cdot m \cdot x^{m-1} + x^m\cdot e^x$

$2^\text{nd}$ Derivative: $e^x\cdot m \cdot (m-1)\cdot x^{m-2} + 2 \cdot e^x\cdot m \cdot x^{m-1} + x^m\cdot e^x$

$3^\text{rd}$ Derivative: $e^x\cdot m \cdot (m-1) \cdot (m-2)\cdot x^{m-3} + 3 \cdot e^x \cdot m \cdot (m-1) \cdot x^{m-2} + 3 \cdot e^x \cdot m \cdot x^{m-1} + \cdot x^m \cdot e^x$

From here how do I calculate the $n^{th}$ derivative?

Thanks. :)

rndflas
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    The $n$ in "nth-derivative" is not the same as in $x^n$, right? – HDE 226868 Nov 23 '14 at 19:25
  • Thanks for pointing that out... I edited the question now.. – rndflas Nov 23 '14 at 19:29
  • I don't believe your third derivative is correct. The third coefficient should be a $3$, and the fourth should be a $1$. With that you should be see a pattern. – Andrey Kaipov Nov 23 '14 at 19:29
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    See the answer to your own question http://math.stackexchange.com/questions/1035189/nth-derivative-of-a-function?rq=1 – Matt Samuel Nov 23 '14 at 19:31
  • @AndreyKaipov you are correct! I still have difficulty on forming the pattern. I mean, I can see what next term would look like, but have no idea on coefficients. – rndflas Nov 23 '14 at 19:39
  • @MattS It has been answered there, but the answers there were with General Leibniz rule, which I couldn't use because, I first need to find the pattern and then use mathematical induction to prove it. But thanks for pointing out. – rndflas Nov 23 '14 at 19:43

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