anonymous
  • anonymous
Find thw anti-derivative of \[f(x)=\frac{ -1 }{ x^2 }e ^{1/x}\]
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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johnweldon1993
  • johnweldon1993
\[\large -\int \frac{e^{1/x}}{x^2}\] Make a u-sub of 1/x
johnweldon1993
  • johnweldon1993
Oh...what have you learned?
anonymous
  • anonymous
How do you write in this format? f(x)=F(x)+C

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zepdrix
  • zepdrix
\[\Large\rm \int\limits f(x)dx=F(x)+C\]Where F(x) is the anti-derivative of f(x). Did you understand the u-substitution?
anonymous
  • anonymous
Is there another way to solve it without using the u-substitution?
freckles
  • freckles
i can't think of a better way
anonymous
  • anonymous
The reason that I ask that was because the question was at the beginning of antiderivatives/integration. I learned the u-substitution as a last topic...
ganeshie8
  • ganeshie8
you may use advanced guessing what is the derivative of \(\large e^{1/x}\) ?
anonymous
  • anonymous
e^1/x
ganeshie8
  • ganeshie8
what do you know about chain rule
anonymous
  • anonymous
Chain rule applies to composite function. It describe of outer evulated at inner, times derivative of inner
johnweldon1993
  • johnweldon1993
So using that logic...if we consider the "inner" as 1/x The derivative of \(\large e^{1/x}\) would be \(\large \frac{d(1/x)}{dx} \times e^{1/x}\) So what is the derivative of \(\large 1/x\) ?
anonymous
  • anonymous
lnx?
freckles
  • freckles
you found the antiderivative for positive x
freckles
  • freckles
that you found the antiderivative of 1/x for positive x the question was to differentiate
ganeshie8
  • ganeshie8
derivative if lnx is 1/x derivative of 1/x is not lnx
anonymous
  • anonymous
ooh... derivative of 1/x is -1/x^2 ...
johnweldon1993
  • johnweldon1993
Correct...and to finish up from the previous post \[\large \frac{d}{dx}e^{1/x} = e^{1/x} \times \frac{d}{dx}\frac{1}{x}\] Since we just found the later \(\large \frac{d}{dx} \frac{1}{x} = -\frac{1}{x^2}\) we have \[\large \frac{d}{dx}e^{1/x} = -\frac{e^{1/x}}{x^2}\] right? Now what do you notice?
anonymous
  • anonymous
So F(x)= e^{1/x} +C ?
johnweldon1993
  • johnweldon1993
Perfect! :)
anonymous
  • anonymous
Thank you! :) I got it now. I was messed up b/w derivatives and anti-derivatives. My bad...
johnweldon1993
  • johnweldon1993
Yeah sometimes they can be hard to keep straight...but as long as you can derive them you're all set!

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