# Thread: how to dolce this?deviative of exponenetial functions

1. ## how to do this?deviative of exponenetial functions

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

the outcome supposs to be like this

$\displaystyle \frac{2x-3}{e^2^x}$

i dont even know which part shld i deviate first

i am trying to learn the steps to solving this pls help me

2. Originally Posted by neodeath
$\displaystyle \frac{x^2 -1}{ e^2^x(x+1)}$

the outcome supposs to be like this

$\displaystyle \frac{2x-3}{e^2^x}$

i dont even know which part shld i deviate first

i am trying to learn the steps to solving this pls help me
$\displaystyle \frac{x^2 -1}{ e^2^x(x+1)}=\frac{(x+1)(x-1)}{e^{2x}(x+1)}=\frac{x-1}{e^{2x}}$

Which can be differentiated using the quotient rule, or if you are like
me and can't be bothered with remembering that, we can rewrite this as:

$\displaystyle \frac{d}{dx}\ \frac{x-1}{e^{2x}}=\frac{d}{dx}(x-1)e^{-2x}$

which can be done using the product rule:

$\displaystyle \frac{d}{dx}(x-1)e^{-2x}=e^{-2x}+(x-1)(-2)e^{-2x}=\frac{3-2x}{e^{2x}}$

RonL

3. wow ddnt know cna use product and quotion rule

by the way i made some mistakes too the ans supposs to be like this:

$\displaystyle -\frac{2x-3}{e^2^x}$

seriosuly embaress >.<

but any way so to do question like this i shld simplify them then deviate them rite?

4. Originally Posted by neodeath
wow ddnt know cna use product and quotion rule

by the way i made some mistakes too the ans supposs to be like this:

$\displaystyle -\frac{2x-3}{e^2^x}$

seriosuly embaress >.<

but any way so to do question like this i shld simplify them then deviate them rite?
Yes (note the answer that I got is the same as the one you are now quoting)

RonL