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I had a homework problem to calculate the derivative of $$\frac{x}{e^x}$$ which I did, sucessfully but then I figured there would be a formula for $n^{th}$ derivative, considering how easy it was to find the first derivative. So I calculated a couple of more derivatives with values of $a$ increasing by one i.e. $x^2, x^3 \ldots$ and found the following formula:

$$f^{'}(x) = \frac{x^{n-1}(n-x)}{e^x}$$

and this works for the second, third and fourth derivative. For example, the second derivative of $\frac{x^2}{e^x} is

$$\begin{align} &f^{'}(x) = \frac{x^{n-1}(n-x)}{e^x} \\ &f^{'}(x) = \frac{x^{2-1}(2-x)}{e^x} \\ &f^{'}(x) = \frac{x(n-x)}{e^x} \end{align}$$

and if you check through W|A, it is correct. To verify that this is formula is correct, I took an arbitrary $n$.

\begin{align} \require{cancel} f^{'}(x) &= \frac{x^{n}}{e^x} \\ &= \frac{nx^{n-1}e^x - x^ne^x}{(e^x)^2} \\ &= \frac{x^{n-1}e^x (n - x)}{e^{2x}} \\ &= \frac{x^{n-1}\cancel{e^x} (n-x)}{\cancel{e^{2x}}} \\ &= \frac{x^{n-1}(n-x)}{e^x} \\ \end{align}

And there you have it. Is this a valid way to show the derivation of the formula. Also, is the formula correct?

Thanks a bunch!

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@Downvoter, care to comment? –  gekkostate Nov 12 '13 at 19:09
    
Hint: I think you may have to re-phrase the question: "Find the first derivative of $x^n/e^x$..." –  Andrea L. Nov 12 '13 at 19:42
    
@AndreaL. I'm looking to find a general formula for the $n^th$ derivative of the that function so isn't my title correct? –  gekkostate Nov 12 '13 at 19:48
    
Sorry, I may have sound rude; but since I want you to "see" the basic concepts at work rather than giving just the result, maybe this tip would help. You derived once a function whose exponent is related to $n$, and as you shown, no matter how it can be derived, the factor $e^x$ simplifies (the denominator is constant). But the numerator now is the function $nx^{n-1}-x^n$, which is only the first derivation of the quotient. You have to carry on deriving $n$ times the numerator, which will lead to a pattern... –  Andrea L. Nov 12 '13 at 20:17
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1 Answer

Hint:
Apply Leibniz rule to the product $x^n e^{-x}.$

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I have updated the question, I believe it was unclear before. I am not looking for the $n^{th}$ derivative of $x^n e^{-x}$ but rather, I'm looking for the first derivative of $x^{a}*e^{-x}$ where $a$ is any positive greater than zero. –  gekkostate Nov 12 '13 at 21:10
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