Ex 12.2

Chapter 12 Class 11 Limits and Derivatives
Serial order wise

### Transcript

Ex 12.2, 4 Find the derivative of the following functions from first principle. (iv) (π₯ + 1)/(π₯ β 1) Let f (x) = (π₯ + 1)/(π₯ β 1) We need to find Derivative of f(x) i.e. fβ (x) We know that fβ(x) = limβ¬(hβ0) fβ‘γ(x + h) β f(x)γ/h Here, f (x) = (π₯ + 1)/(π₯ β 1) So, f (x + h) = ((π₯ + β) + 1)/((π₯ + β) β 1) Ex 12.2, 4 Find the derivative of the following functions from first principle. (iv) (π₯ + 1)/(π₯ β 1) Let f (x) = (π₯ + 1)/(π₯ β 1) We need to find Derivative of f(x) i.e. fβ (x) We know that fβ(x) = limβ¬(hβ0) fβ‘γ(x + h) β f(x)γ/h Here, f (x) = (π₯ + 1)/(π₯ β 1) So, f (x + h) = ((π₯ + β) + 1)/((π₯ + β) β 1) Putting values fβ(x) = limβ¬(hβ0)β‘γ([((π₯ + β) + 1)/(π₯ + β β 1)] β[ (π₯ + 1)/(π₯ β 1)])/hγ = limβ¬(hβ0)β‘γ((π₯ + β + 1)/(π₯ + β β 1) β (π₯ + 1)/(π₯ β 1))/hγ = limβ¬(hβ0)β‘γ(((π₯ β 1)(π₯ + β + 1) β (π₯ + 1)( π₯ + β β 1))/(( π₯ + β β 1 ) (π₯ β 1)))/βγ = limβ¬(hβ0)β‘γ((π₯ β 1) ((π₯ + 1) + β) β (π₯ + 1)( (π₯ β 1) + β))/(β( π₯ + β β 1 ) (π₯ β 1))γ = limβ¬(hβ0)β‘γ((π₯ β 1)(π₯ + 1) + (π₯ β1)β β (π₯ + 1)(π₯ β 1) β (π₯ + 1) β)/(β( π₯ + β β 1 ) (π₯ β 1))γ = limβ¬(hβ0)β‘γ((π₯2 β 1) + π₯β β (π₯2 β 1) β π₯β β β)/(β( π₯ + β β 1 ) (π₯ β 1))γ = (πππ)β¬(ββ0)β‘γ(β 2β )/(β (π₯ + β β 1) (π₯ β 1))γ = limβ¬(hβ0)β‘γ(β2)/((π₯ + β β 1) (π₯ β 1))γ Putting h = 0 = (β2)/((π₯ + 0 β 1)(π₯ β 1)) = (β2)/((π₯ β 1) (π₯ β 1)) = (β2)/(π₯ β 1)^2 Hence, fβ(x) = (βπ)/(π β π)^π