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.
The text passage contains a mathematical question and an answer to that question, specifically regarding the calculation of the second derivative of a function. The question is posed by a user who is stuck on the problem, and the answer is provided by another user, with further discussion and simplification.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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.
user
The text passage contains a mathematical question and an answer to that question, specifically regarding the calculation of the second derivative of a function. The question is posed by a user who is stuck on the problem, and the answer is provided by another user, with further discussion and simplification.

TYPE: SINGLE_QUESTION_WITH_ANSWER