## Calculus 10th Edition

$f'(x)=-\dfrac{13}{(2x-5)^2}, f'(9)=-\dfrac{1}{13}.$
Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=x+4; u'(x)=1$ $v(x)=2x-5; v'(x)=2$ $f'(x)=\dfrac{1(2x-5)-2(x+4)}{(2x-5)^2}=-\dfrac{13}{(2x-5)^2}.$ $f'(9)=-\dfrac{13}{(2(9)-5)^2}=-\dfrac{1}{13}.$ A graphing utility was used to verify this result.