Ex 12.2
Ex 12.2, 2
Ex 12.2, 3
Ex 12.2, 4 (i) Important
Ex 12.2, 4 (ii)
Ex 12.2, 4 (iii) Important You are here
Ex 12.2, 4 (iv)
Ex 12.2, 5
Ex 12.2, 6
Ex 12.2, 7 (i) Important
Ex 12.2, 7 (ii)
Ex 12.2, 7 (iii) Important
Ex 12.2, 8
Ex 12.2, 9 (i)
Ex 12.2, 9 (ii) Important
Ex 12.2, 9 (iii)
Ex 12.2, 9 (iv) Important
Ex 12.2, 9 (v)
Ex 12.2, 9 (vi)
Ex 12.2, 10 Important
Ex 12.2, 11 (i)
Ex 12.2, 11 (ii) Important
Ex 12.2, 11 (iii) Important
Ex 12.2, 11 (iv)
Ex 12.2, 11 (v) Important
Ex 12.2, 11 (vi)
Ex 12.2, 11 (vii) Important
Last updated at May 7, 2024 by Teachoo
Ex 12.2, 4 Find the derivative of the following functions from first principle. (iii) 1/๐ฅ^2 Let f (x) = 1/๐ฅ^2 We need to find derivative of f(x) i.e. fโ (x) We know that fโ(x) = (๐๐๐)โฌ(โโ0) ๐โกใ(๐ฅ + โ) โ ๐(๐ฅ)ใ/โ Here, f (x) = 1/๐ฅ^2 So, f (x + h) = 1/ใ(๐ฅ + โ)ใ^2 Putting values fโ(x) = (๐๐๐)โฌ(โโ0)โกใ(1/ใ(๐ฅ + โ)ใ^2 โ 1/๐ฅ^2 )/โใ = (๐๐๐)โฌ(โโ0)โกใ(ใใ๐ฅ ใ^2 โ (๐ฅ + โ)ใ^2/(ใ(๐ฅ + โ)ใ^2 ๐ฅ^2 ))/โใ = (๐๐๐)โฌ(โโ0)โกใ( ๐ฅ2 โ (๐ฅ + โ)2)/(โ๐ฅ2 (๐ฅ + โ)2)ใ = (๐๐๐)โฌ(โโ0)โกใ((๐ฅ โ ( ๐ฅ + โ )) (๐ฅ + (๐ฅ + โ)))/(โ๐ฅ2 (๐ฅ + โ)2)ใ = (๐๐๐)โฌ(โโ0)โกใ( (๐ฅ โ๐ฅ โ โ) (๐ฅ + ๐ฅ + โ))/(โ.๐ฅ2 (๐ฅ + โ)2)ใ (As a2 โ b2 = (a โ b) (a + b)) = (๐๐๐)โฌ(โโ0)โกใ((โโ)(2๐ฅ + โ))/(โ๐ฅ2 (๐ฅ + โ)2)ใ = (๐๐๐)โฌ(โโ0)โกใ((โ1) 2๐ฅ + โ)/(๐ฅ2 (๐ฅ + โ)2)ใ Putting h = 0 = ((โ1) 2๐ฅ + 0)/(๐ฅ2 (๐ฅ + 0)2) = ((โ1) 2๐ฅ)/(๐ฅ2 (๐ฅ)2) = (โ2๐ฅ)/๐ฅ4 = (โ2)/๐ฅ^3 Thus, fโ(x) = (โ๐)/๐^๐