$\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)}
$
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
$\displaystyleOriginally 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
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?