# [SOLVED] Tricky Diff question

• Mar 7th 2010, 09:18 AM
dojo
[SOLVED] Tricky Diff question
Hi - I just can't seem to get this one right. Looking to get the first derivative...

(1 + xe^x)/(1 - xe^x)

I assume that you use the quotient rule but I end up in a mess.

Could someone please talk me thorugh this one?

Thanks, D
• Mar 7th 2010, 09:29 AM
Hello dojo
Quote:

Originally Posted by dojo
Hi - I just can't seem to get this one right. Looking to get the first derivative...

(1 + xe^x)/(1 - xe^x)

I assume that you use the quotient rule but I end up in a mess.

Could someone please talk me thorugh this one?

Thanks, D

Yes, you'll need to use both the quotient and the product rules:
$\displaystyle f(x) = \frac{1+xe^x}{1-xe^x}$

$\displaystyle \Rightarrow f'(x)= \frac{(1-xe^x)(xe^x+e^x)-(1+xe^x)(-xe^x-e^x)}{(1-xe^x)^2}$
$\displaystyle = \frac{e^x(1-xe^x)(x+1)+e^x(1+xe^x)(x+1)}{(1-xe^x)^2}$

$\displaystyle = \frac{e^x(x+1)(1-xe^x+1+xe^x)}{(1-xe^x)^2}$

$\displaystyle = \frac{2e^x(x+1)}{(1-xe^x)^2}$

• Mar 8th 2010, 04:37 AM
dojo
The book gives the answer with the same numerator as you have bu the denom as (xe^2 - 1)^2

Thanks for the workings - I think my algebra needs a bit of work. I think the book must be wrong as the quotien rule does state that v^2 is the denom.

Thanks
• Mar 8th 2010, 06:00 AM
Hello dojo
Quote:

Originally Posted by dojo
The book gives the answer with the same numerator as you have bu the denom as (xe^2 - 1)^2

Thanks for the workings - I think my algebra needs a bit of work. I think the book must be wrong as the quotien rule does state that v^2 is the denom.

Thanks

Since $\displaystyle (a-b)^2$ is the same as $\displaystyle (b-a)^2$, you could certainly write the denominator as $\displaystyle (xe^x-1)^2$, but not $\displaystyle (xe^2-1)^2$.

Are you sure that's what the book said? If so, it's just a typo.