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We recently had a test and on the test we were asked to make a formula for the $n^{th}$ derivative of $e^{2x}$. My friends and I developed different formulas for the question. I developed $\left(\frac{1}{2}\right)^{-n}*e^{2x}$ where $n$ is the $n^{th}$ derivative and my friend developed $(2^n)*e^{2x}$ where $n$ again is the $n^{th}$ derivative. I boldly claimed that there could be countless solutions to the problem, I believe I am wrong (without reason). How can it be proven that there are countless solutions, if there are and if there are not countless solutions then can we count how many there are?

Thanks

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Your solution and your friend’s are the same: there are infinitely many different ways to express the solution, but there is only one solution. –  Brian M. Scott Apr 17 '13 at 10:41
    
@BrianM.Scott hmmm okay. But I have another question then, if there are infinitely many ways to arrive to the solution, how can this be proven? (Sorry if the question is naive). –  Jeel Shah Apr 17 '13 at 10:44
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Not infinitely many ways to arrive at it, but infinitely many ways to write it. Take any expression $u$ that evaluates to $2$, for instance, and write $u^ne^{2x}$. For instance, $u$ could be $2+k-k)$ for any $k$. Or you could replace $2^n$ by $b^{n\log_b2}$ for any $b>1$. –  Brian M. Scott Apr 17 '13 at 10:47

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Since $(\frac{1}{2})^{-n}e^{2x}=\frac{1}{2^{-n}}e^{2x}=2^ne^{2x}$ so both you and your friend have the same answer. There is only one solution in any such problem of the $n$th derivative.

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