## anonymous one year ago Find thw anti-derivative of $f(x)=\frac{ -1 }{ x^2 }e ^{1/x}$

1. johnweldon1993

$\large -\int \frac{e^{1/x}}{x^2}$ Make a u-sub of 1/x

2. johnweldon1993

Oh...what have you learned?

3. anonymous

How do you write in this format? f(x)=F(x)+C

4. 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?

5. anonymous

Is there another way to solve it without using the u-substitution?

6. freckles

i can't think of a better way

7. 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...

8. ganeshie8

you may use advanced guessing what is the derivative of $$\large e^{1/x}$$ ?

9. anonymous

e^1/x

10. ganeshie8

what do you know about chain rule

11. anonymous

Chain rule applies to composite function. It describe of outer evulated at inner, times derivative of inner

12. 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 $$\large 1/x$$ ?

13. anonymous

lnx?

14. freckles

you found the antiderivative for positive x

15. freckles

that you found the antiderivative of 1/x for positive x the question was to differentiate

16. ganeshie8

derivative if lnx is 1/x derivative of 1/x is not lnx

17. anonymous

ooh... derivative of 1/x is -1/x^2 ...

18. 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?

19. anonymous

So F(x)= e^{1/x} +C ?

20. johnweldon1993

Perfect! :)

21. anonymous

Thank you! :) I got it now. I was messed up b/w derivatives and anti-derivatives. My bad...

22. johnweldon1993

Yeah sometimes they can be hard to keep straight...but as long as you can derive them you're all set!