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Without using Mathematical Induction to calculus the n'th derivative of the following function.

$y_{n}=x^{n-1}e^{1/x}$ , $n\in\mathbb{N}$

Find : $\frac{d^n}{dx^n}y_n$

I tried to finish the question from the given answer : $ y^{(n)}_{n}=\frac{(-1)^{n}e^{1/x}}{x^{n+1}} $

So, I want to know other method(s) to get the answer.

Thanks for your help!! :)

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Why would you not want to use induction? The $n$-th derivative is itself recursively defined, so it seems a little silly to avoid induction. –  Pedro Tamaroff Sep 20 '12 at 17:21
    
My induction proof comes from the given answer : $ y^{(n)}_{n}=\frac{(-1)^{n}e^{1/x}}{x^{n+1}} $ So, I want to know other method(s) to get the answer. –  cwk709394 Sep 20 '12 at 17:25
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1 Answer 1

up vote 1 down vote accepted

Begin with

$${y_n}(x) = {x^{n - 1}}{e^{1/x}}$$

Then, taking the first derivative, we get

$${y_n}'(x) = \left( {n - 1} \right){x^{n - 2}}{e^{1/x}} - {x^{n - 3}}{e^{1/x}}$$

This is

$$\tag{1}{y_n}'(x) = \left( {n - 1} \right){y_{n - 1}(x)} - {y_{n - 2}}(x) $$

You then get

$${y_n}''(x) = \left( {n - 1} \right)y{'_{n - 1}}(x) - y{'_{n - 2}}(x)$$

so you have to use $(1)$ to obtain that exclusively in terms of $y_n$.

You can do this for a couple of terms, conjecture a general formula, and prove it by induction. In fact, the formula will be only in terms of $n!/(n-k)!$ and $y_{n-k}$, it seems. When you get a general formula for $y_n^{(k)}$, you can plug in $n$ and get the formula you want.

Another option would be

$$\eqalign{ & {y_n}' = \left( {n - 1} \right)\frac{1}{x}{x^{n - 1}}{e^{1/x}} - \frac{1}{{{x^2}}}{x^{n - 1}}{e^{1/x}} \cr & {y_n}' = \left( {n - 1} \right)\frac{1}{x}{y_n} - \frac{1}{{{x^2}}}{y_n} \cr & {y_n}^\prime (x) = \left( {n - 1} \right)\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right){y_n} \cr} $$

Then

$${y_n}''(x) = \left( {n - 1} \right)\left( { - \frac{1}{{{x^2}}} + \frac{2}{{{x^3}}}} \right){y_n} + \left( {n - 1} \right)\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)y{'_n}$$

So

$${y_n}''(x) = \left( {n - 1} \right)\left( { - \frac{1}{{{x^2}}} + \frac{2}{{{x^3}}}} \right){y_n} + {\left( {n - 1} \right)^2}{\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)^2}{y_n}$$

But it seems much more complicated, due to the repeated use of the product rule and chain rule involved.

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Thanks Peter!! :) –  cwk709394 Sep 20 '12 at 17:36
    
Welcome. I think this is still note very optimal, so I'll edit with something better if I can. –  Pedro Tamaroff Sep 20 '12 at 17:57
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