text: 1. ## Second Derivative Help
  
  I am stuck on a problem concerning second derivatives and was wondering if you could provide
  some help.
  
  orignial function:
  
  f(x)= (e^x)/x
  
  I believe i figured out the first derivative using the quotient rule.
  
  f'(x)= [e^x(x-1)] / x^2
  
  Now, I am not sure what to do with the top. I know the chain rule is
  somewhere in there. Any advice would be great, thank you.
  
  2. Originally Posted by lostsheep
  I am stuck on a problem concerning second derivatives and was wondering if you could provide
  some help.
  
  orignial function:
  
  f(x)= (e^x)/x
  
  I believe i figured out the first derivative using the quotient rule.
  
  f'(x)= [e^x(x-1)] / x^2
  
  Now, I am not sure what to do with the top. I know the chain rule is
  somewhere in there. Any advice would be great, thank you.
  the product rule actually:
  
  $f'(x) = \frac {e^x ( x - 1)}{x^2}$
  
  By the quotient rule, we get:
  
  $f''(x) = \frac { \left( x^2 \right)( e^x (x - 1))' - (e^x(x - 1)) \left( x^2 \right)'}{x^4}$
  
  $= \frac {x^2[e^x(x - 1) + e^x] - 2x(e^x(x - 1))}{x^4}$
  
  and simplify
  
  and that's using the quotient rule. were it me though, i'd change the function so i can use the product rule all the way, i find the quotient rule very annoying
  
  3. I think you may separate it into two parts, as
  
  f''(x)=x(e^x)/(x^2) d/dx - (e^x)/(x^2) d/dx
  =(e^x)/x d/dx - (e^x)/(x^2) d/dx
  =(e^x(x-1))/(x^2) - ((x^2*e^x)-(e^x(4*x^3))/(x^4)
  ...
  
  and simplify it...
  
  I hope it is right..
  
  4. Originally Posted by Jhevon
  the product rule actually:
  
  $f'(x) = \frac {e^x ( x - 1)}{x^2}$
  
  By the quotient rule, we get:
  
  $f''(x) = \frac { \left( x^2 \right)( e^x (x - 1))' - (e^x(x - 1)) \left( x^2 \right)'}{x^4}$
  
  $= \frac {x^2[e^x(x - 1) + e^x] - 2x(e^x(x - 1))}{x^4}$
  
  and simplify
  
  and that's using the quotient rule. were it me though, i'd change the function so i can use the product rule all the way, i find the quotient rule very annoying
  
  How would you change the function to use only the product rule?
  
  As for simplifying, I got
  
  [e^x(x-1+1)(-2x^2 -2x) ] / x^2
  
  Is that even remotely close? Thanks for all the help.
  
  5. Originally Posted by lostsheep
  How would you change the function to use only the product rule?
  like this:
  
  $f(x) = \frac {e^x}{x}$
  
  $\Rightarrow f(x) = x^{-1}e^x$
  
  $\Rightarrow f'(x) = -x^{-2}e^x + x^{-1}e^x$
  
  $\Rightarrow f''(x) = 2x^{-3}e^x + -x^{-2}e^x - x^{-2}e^x + x^{-1}e^x$
  
  $= e^x \left( 2x^{-3} - 2x^{-2} + x^{-1} \right)$
  
  and simplify that. easy huh?
  
  As for simplifying, I got
  
  [e^x(x-1+1)(-2x^2 -2x) ] / x^2
  
  Is that even remotely close? Thanks for all the help.
  that is not simplified. for one obvious reason, we could write x - 1 + 1 as just x. there are other ways you can simplify
  
  6. Originally Posted by Jhevon
  like this:
  
  $f(x) = \frac {e^x}{x}$
  
  $\Rightarrow f(x) = x^{-1}e^x$
  
  $\Rightarrow f'(x) = -x^{-2}e^x + x^{-1}e^x$
  
  $\Rightarrow f''(x) = 2x^{-3}e^x + -x^{-2}e^x - x^{-2}e^x + x^{-1}e^x$
  
  $= e^x \left( 2x^{-3} - 2x^{-2} + x^{-1} \right)$
  
  and simplify that. easy huh?
  
  that is not simplified. for one obvious reason, we could write x - 1 + 1 as just x. there are other ways you can simplify
  
  That is very interesting, i suppose you can just elminate the quotiet rule altogether.
  
  Simplifing is one of my weak points.
  
  so thats say I did use the product rule.
  
  f''(x)= e^x (2x^-3 -2x^-2 + x^-1)
  
  then simplified would be;
  
  f''(x) = xe^x ( 2x^-2 -2x^-1 -1)
  
  f''(x) = xe^x / (2x^2 -2x -1 )
  
  If that is not corrected, could you provide it simplified in using fraction/ division. I want to realize my mistakes. Thanks a lot.
  
  7. Originally Posted by lostsheep
  That is very interesting, i suppose you can just elminate the quotiet rule altogether.
  
  Simplifing is one of my weak points.
  
  so thats say I did use the product rule.
  
  f''(x)= e^x (2x^-3 -2x^-2 + x^-1)
  
  then simplified would be;
  
  f''(x) = xe^x ( 2x^-2 -2x^-1 -1)
  
  f''(x) = xe^x / (2x^2 -2x -1 )
  
  If that is not corrected, could you provide it simplified in using fraction/ division. I want to realize my mistakes. Thanks a lot.
  $e^x \left( 2x^{-3} - 2x^{-2} + x^{-1} \right)$
  
  $= e^x \left( \frac{2}{x^3} - \frac {2}{x^2} + \frac {1}{x} \right)$
  
  $= e^x \left( \frac{x^2 - 2x + 2}{x^3} \right)$
  
  you factored out an x incorrectly, so you were wrong from the beginning. you could have factored out the x^{-3} and you would have gotten what i did almost immediately. try it
  
  8. Originally Posted by Jhevon
  $e^x \left( 2x^{-3} - 2x^{-2} + x^{-1} \right)$
  
  $= e^x \left( \frac{2}{x^3} - \frac {2}{x^2} + \frac {1}{x} \right)$
  
  $= e^x \left( \frac{x^2 - 2x + 2}{x^3} \right)$
  
  you factored out an x incorrectly, so you were wrong from the beginning. you could have factored out the x^{-3} and you would have gotten what i did almost immediately. try it
  Thank you very much. I need to work on that. Natural Logs and e just throw me off when it comes to this, as well as the simplifing. Thanks again.