Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
Your English was perfect, so I removed your apology for it ;) –  Sharkos May 26 '13 at 23:41

2 Answers 2

up vote 5 down vote accepted

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:

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

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.