## Calculus: Early Transcendentals 8th Edition

$f'(x)=\dfrac{ad-bc}{(cx+d)^{2}}$
$f(x)=\dfrac{ax+b}{cx+d}$ (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: $...=\dfrac{acx+ad-acx-bc}{(cx+d)^{2}}=\dfrac{ad-bc}{(cx+d)^{2}}$