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Ex 12.2

Ex 12.2, 1

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Ex 12.2, 4 (i) Important You are here

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Ex 12.2, 4 (iii) Important

Ex 12.2, 4 (iv)

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Ex 12.2, 7 (i) Important

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Ex 12.2, 9 (i)

Ex 12.2, 9 (ii) Important

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Ex 12.2, 9 (iv) Important

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Ex 12.2, 10 Important

Ex 12.2, 11 (i)

Ex 12.2, 11 (ii) Important

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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 29, 2023 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