# Ex 13.2, 4 (i) - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

Last updated at Aug. 28, 2021 by Teachoo

Ex 13.2

Ex 13.2, 1

Ex 13.2, 2

Ex 13.2, 3

Ex 13.2, 4 (i) Important You are here

Ex 13.2, 4 (ii)

Ex 13.2, 4 (iii) Important

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 Aug. 28, 2021 by Teachoo

Ex 13.2, 4 Find the derivative of the following functions from first principle. (i) x3 – 27 Let f(x) = x3 – 27 We need to find Derivative of f(x) i.e. f’ (x) We know that f’(x) = lim┬(h→0) f〖(x + h) − f(x)〗/h f (x) = x3 – 27 f (x + h) = (x + h)3 – 27 Putting values f’(x) = lim┬(h→0)〖(((x + h)3 − 27) − (x3 − 27))/h〗 = lim┬(h→0)〖((x + h)3 − 27− x3 + 27)/h〗 = lim┬(h→0)〖((x + h)3 − x3 − 27 + 27)/h〗 = lim┬(h→0)〖((x + h)3 − x3 )/h〗 = (𝑙𝑖𝑚)┬(ℎ→0)〖(𝑥3 + ℎ3 + 3𝑥2 ℎ + 3𝑥ℎ2 − 𝑥3)/ℎ〗 = (𝑙𝑖𝑚)┬(ℎ→0)〖(ℎ3 + 3𝑥2 ℎ + 3𝑥ℎ2 − 𝑥3 + 𝑥3)/ℎ〗 = (𝑙𝑖𝑚)┬(ℎ→0)〖(ℎ ( ℎ2 +3𝑥2 + 3𝑥ℎ) )/ℎ〗 = lim┬(h→0)〖ℎ2+3𝑥2+3𝑥ℎ〗 Putting h = 0 = (0)2 + 3x2 + 3x(0) = 0 + 3x2 + 0 = 3x2 Hence, f’(x) = 3x2