Thread: Derivative help

$\displaystyle D_r (\frac{2r+3}{3r-2})$

Originally Posted by ^_^Engineer_Adam^_^

$\displaystyle D_r (\frac{2r+3}{3r-2})$

You need the quotient rule:

$\displaystyle D_r \frac{f(r)}{g(r)}=\frac{g(r)f'(r)-f'(r)g(r)}{[g(r)]^2}$

here:

$\displaystyle f(r)=2r+3$,

$\displaystyle f'(r)=2$,

$\displaystyle g(r)=3r-2$,

$\displaystyle g'(r)=3$

so:

$\displaystyle D_r \left( \frac{2r+3}{3r-2} \right)=\frac{g(r)f'(r)-f(r)g'(r)}{[g(r)]^2}
=\frac{2(3r-2)-3(2r+3)}{(3r-2)^2}=-\frac{13}{(3r-2)^2}
$

RonL

Originally Posted by CaptainBlank

so:

$\displaystyle D_r \left( \frac{2r+3}{3r-2} \right)=\frac{g(r)f'(r)-f(r)g'(r)}{[g(r)]^2}
=\frac{2(3r-2)-3(2r+3)}{(3r-2)^2}=-\frac{13}{(3r-2)^2}
$

I see you learned the quotient rule!

Originally Posted by ThePerfectHacker

I see you learned the quotient rule!

I'm afraid not; I derived it.

RonL