Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises - Page 189: 30

$f'(x)=\dfrac{ad-bc}{(cx+d)^{2}}$

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