

Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month
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 Sept. 6, 2021 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