Answer
$f'(x)=\dfrac{ad-bc}{(cx+d)^{2}}$
Work Step by Step
$f(x)=\dfrac{ax+b}{cx+d}$ (Here, $a$, $b$, $c$ and $d$ are constants)
Differentiate using the quotient rule:
$f'(x)=\dfrac{(cx+d)(ax+b)'-(ax+b)(cx+d)'}{(cx+d)^{2}}=...$
$...=\dfrac{(cx+d)(a)-(ax+b)(c)}{(cx+d)^{2}}$
Simplify:
$...=\dfrac{acx+ad-acx-bc}{(cx+d)^{2}}=\dfrac{ad-bc}{(cx+d)^{2}}$
