Question

`(ax + b)/(cx + d)`

Answer 1

Let `f(x) = (ax + b)/(cx + d)`  ......(i)

⇒ `f(x + Δx) = (a(x + Δx) + b)/(c(x + Δx) + d)`  .....(ii)

Subtracting equation (i) from equation (ii) we get

`f(x + Δx) - f(x) = (a(x + Δx) + b)/(c(x + Δx) + d) - (ax + b)/(cx + d)`

Dividing both sides by Δx and take the limit, we get

`lim_(Δx -> 0) (f(x + Δx) - f(x))/(Δx) = lim_(Δx -> 0) ((a(x + Δx) + b)/(c(x + Δx) + d) - (ax + b)/(cx + d))/(Δx)`

⇒ f'(x) = `lim_(Δx -> 0) ((ax + aΔx + b)(cx + d) - (ax + b)(cx + cΔx + d))/([c(x + Δx) + d](cx + d) * Δx)`  ......[Using definition of differentiation]

`acx^2 + acΔx * x + bcx  + adx + adΔx + bd`

= `lim_(Δx -> 0) (-acx^2 - acΔx * x - adx - bcx - bc * Δx - bd)/((cx + cΔx + d)(cx + d) * Δx)`

= `lim_(Δx -> 0) ((ad - bc)Δx)/((cx + c*Δx + d)(cx + d))`

Taking limit, we have

= `((ad - bc))/((cx + d)(cx + d))`

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