Ex 13.2

Chapter 13 Class 11 Limits and Derivatives
Serial order wise

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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) = (πππ)β¬(ββ0) πβ‘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) = (πππ)β¬(ββ0)β‘γ(1/γ(π₯ + β)γ^2 β 1/π₯^2 )/βγ = (πππ)β¬(ββ0)β‘γ(γγπ₯ γ^2 β (π₯ + β)γ^2/(γ(π₯ + β)γ^2 π₯^2 ))/βγ = (πππ)β¬(ββ0)β‘γ( π₯2 β (π₯ + β)2)/(βπ₯2 (π₯ + β)2)γ = (πππ)β¬(ββ0)β‘γ((π₯ β ( π₯ + β )) (π₯ + (π₯ + β)))/(βπ₯2 (π₯ + β)2)γ = (πππ)β¬(ββ0)β‘γ( (π₯ βπ₯ β β) (π₯ + π₯ + β))/(β.π₯2 (π₯ + β)2)γ (As a2 β b2 = (a β b) (a + b)) = (πππ)β¬(ββ0)β‘γ((ββ)(2π₯ + β))/(βπ₯2 (π₯ + β)2)γ = (πππ)β¬(ββ0)β‘γ((β1) 2π₯ + β)/(π₯2 (π₯ + β)2)γ Putting h = 0 = ((β1) 2π₯ + 0)/(π₯2 (π₯ + 0)2) = ((β1) 2π₯)/(π₯2 (π₯)2) = (β2π₯)/π₯4 = (β2)/π₯^3 Thus, fβ(x) = (βπ)/π^π