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Ex 13.2 (Term 2)
Ex 13.2, 2
Ex 13.2, 3
Ex 13.2, 4 (i) Important
Ex 13.2, 4 (ii)
Ex 13.2, 4 (iii) Important
Ex 13.2, 4 (iv) You are here
Ex 13.2, 5
Ex 13.2, 6
Ex 13.2, 7 (i) Important
Ex 13.2, 7 (ii)
Ex 13.2, 7 (iii) Important
Ex 13.2, 8
Ex 13.2, 9 (i)
Ex 13.2, 9 (ii) Important
Ex 13.2, 9 (iii)
Ex 13.2, 9 (iv) Important
Ex 13.2, 9 (v)
Ex 13.2, 9 (vi)
Ex 13.2, 10 Important
Ex 13.2, 11 (i)
Ex 13.2, 11 (ii) Important
Ex 13.2, 11 (iii) Important
Ex 13.2, 11 (iv)
Ex 13.2, 11 (v) Important
Ex 13.2, 11 (vi)
Ex 13.2, 11 (vii) Important
Last updated at Aug. 28, 2021 by Teachoo
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Ex 13.2, 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¬(hβ0) 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 13.2, 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¬(hβ0) 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¬(hβ0)β‘γ([((π₯ + β) + 1)/(π₯ + β β 1)] β[ (π₯ + 1)/(π₯ β 1)])/hγ = limβ¬(hβ0)β‘γ((π₯ + β + 1)/(π₯ + β β 1) β (π₯ + 1)/(π₯ β 1))/hγ = limβ¬(hβ0)β‘γ(((π₯ β 1)(π₯ + β + 1) β (π₯ + 1)( π₯ + β β 1))/(( π₯ + β β 1 ) (π₯ β 1)))/βγ = limβ¬(hβ0)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¬(hβ0)β‘γ((π₯ β 1)(π₯ + 1) + (π₯ β1)β β (π₯ + 1)(π₯ β 1) β (π₯ + 1) β)/(β( π₯ + β β 1 ) (π₯ β 1))γ = limβ¬(hβ0)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 = (πππ)β¬(ββ0)β‘γ(β 2β )/(β (π₯ + β β 1) (π₯ β 1))γ = limβ¬(hβ0)β‘γ(β2)/((π₯ + β β 1) (π₯ β 1))γ Putting h = 0 = (β2)/((π₯ + 0 β 1)(π₯ β 1)) = (β2)/((π₯ β 1) (π₯ β 1)) = (β2)/(π₯ β 1)^2 Hence, fβ(x) = (βπ)/(π β π)^π