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