$\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

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- Feb 19th 2006, 07:22 AMneodeathhow 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 - Feb 19th 2006, 09:11 AMCaptainBlackQuote:

Originally Posted by**neodeath**

\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 - Feb 19th 2006, 10:09 AMneodeath
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? - Feb 19th 2006, 10:20 AMCaptainBlackQuote:

Originally Posted by**neodeath**

RonL