# Differentiating $y= xe^{1\over x}$

Can someone please help me? I'm trying, but I really can't find the second derivative of $y= xe^{1/x}$. Thanks!

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Your English was perfect, so I removed your apology for it ;) –  Sharkos May 26 '13 at 23:41

By the product rule:

$$y'=x'e^{1/x}+x\left(e^{1/x}\right)'=e^{1/x}-\frac{1}{x}e^{1/x}$$

I used: $y=e^{f(x)}$, then $y'=f'(x)e^{f(x)}$. Then:

$$y''=-\frac{1}{x^2}e^{1/x}+\frac{1}{x^2}e^{1/x}+\frac{1}{x^3}e^{1/x}=\frac{1}{x^3}e^{1/x}$$

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Thi is a hint to solve this kind of problem: you take the logarithm of the function and you have: $$\log y= \log(xe^{\frac{1}{x}})=\frac{1}{x}+\log x$$ so derive it: $$\frac{y'}{y}=-\frac{1}{x^2}+\frac{1}{x}$$ Thus $$y'=y(-\frac{1}{x^2}+\frac{1}{x})=-\frac{e^{\frac{1}{x}}}{x}+e^{\frac{1}{x}}$$

Deriving again with chain rule : $$y''=-\frac{e^{\frac{1}{x}}} {x^2}+ \frac{e^{\frac{1}{x}}}{x^2}+\frac{e^{\frac{1}{x}}}{x^3}= \frac{e^{\frac{1}{x}}}{x^3}$$

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