# how to dolce this?deviative of exponenetial functions

• Feb 19th 2006, 08:22 AM
neodeath
how to do this?deviative of exponenetial functions
$
\frac{x^2 -1}{ e^2^x(x+1)}

$

the outcome supposs to be like this

$
\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
• Feb 19th 2006, 10:11 AM
CaptainBlack
Quote:

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

$

the outcome supposs to be like this

$
\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

$
\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:

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

which can be done using the product rule:

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

RonL
• Feb 19th 2006, 11:09 AM
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:

$
-\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?
• Feb 19th 2006, 11:20 AM
CaptainBlack
Quote:

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:

$
-\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