Ex 12.2

Ex 12.2, 1

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Ex 12.2, 3

Ex 12.2, 4 (i) Important You are here

Ex 12.2, 4 (ii)

Ex 12.2, 4 (iii) Important

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. (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