How do you differentiate $f(x)=(ax+b)/(cx+d)$ ?

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Answer 1 · 2016-11-19T21:04:34

$f'(x)=(ad-bc)/(cx+d)^2$

The Quotient Rule:
$(f/g)^' =(f' cdot g-f cdot g')/(g^2),\ g !=0$

$f(x)=(ax+b)/(cx+d)$
$f'(x)=((ax+b)'(cx+d)-(ax+b)(cx+d)')/(cx+d)^2=(a(cx+d)-(ax+b)c)/(cx+d)^2=(ad-bc)/(cx+d)^2$

How do you differentiate f(x)=(ax+b)/(cx+d)f(x)=(ax+b)/(cx+d)?