Let f (x) = (px + q) (

r

x

+s) ... (i)
Differentiating (i) with respect to x, we get

d

dx

(f (x)) = (px + q)' (

r

x

+s) + (px + q) (

r

x

+s)′
= (p + 0) (

r

x

+s) + (px + q) (

xr′−rx′

x2

+0) = p (

r

x

+s) + (px + q) (

0−r

x2

)
= p (

r

x

+s) -

(px+q)r

x2

=

pr

x

+ps−

pr

x

−

qr

x2

= ps -

qr

x2