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

Last updated at Aug. 28, 2021 by

Ex 13.2, 4 - Find derivative from first principle (i) x3 - 27

Ex 13.2, 4 - Chapter 13 Class 11 Limits and Derivatives - Part 2
Ex 13.2, 4 - Chapter 13 Class 11 Limits and Derivatives - Part 3

Transcript

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

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