Question

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

Answer 1

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

Explanation:

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`