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Ex 13.2
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 You are here
Ex 13.2, 4 (iv)
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 March 30, 2023 by Teachoo
Ex 13.2, 4 Find the derivative of the following functions from first principle. (iii) 1/π₯^2 Let f (x) = 1/π₯^2 We need to find derivative of f(x) i.e. fβ (x) We know that fβ(x) = (πππ)β¬(ββ0) πβ‘γ(π₯ + β) β π(π₯)γ/β Here, f (x) = 1/π₯^2 So, f (x + h) = 1/γ(π₯ + β)γ^2 Putting values fβ(x) = (πππ)β¬(ββ0)β‘γ(1/γ(π₯ + β)γ^2 β 1/π₯^2 )/βγ = (πππ)β¬(ββ0)β‘γ(γγπ₯ γ^2 β (π₯ + β)γ^2/(γ(π₯ + β)γ^2 π₯^2 ))/βγ = (πππ)β¬(ββ0)β‘γ( π₯2 β (π₯ + β)2)/(βπ₯2 (π₯ + β)2)γ = (πππ)β¬(ββ0)β‘γ((π₯ β ( π₯ + β )) (π₯ + (π₯ + β)))/(βπ₯2 (π₯ + β)2)γ = (πππ)β¬(ββ0)β‘γ( (π₯ βπ₯ β β) (π₯ + π₯ + β))/(β.π₯2 (π₯ + β)2)γ (As a2 β b2 = (a β b) (a + b)) = (πππ)β¬(ββ0)β‘γ((ββ)(2π₯ + β))/(βπ₯2 (π₯ + β)2)γ = (πππ)β¬(ββ0)β‘γ((β1) 2π₯ + β)/(π₯2 (π₯ + β)2)γ Putting h = 0 = ((β1) 2π₯ + 0)/(π₯2 (π₯ + 0)2) = ((β1) 2π₯)/(π₯2 (π₯)2) = (β2π₯)/π₯4 = (β2)/π₯^3 Thus, fβ(x) = (βπ)/π^π