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 (Term 2)
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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)
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Ex 13.2, 7 (i) Important
Ex 13.2, 7 (ii)
Ex 13.2, 7 (iii) Important
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Ex 13.2, 9 (i)
Ex 13.2, 9 (ii) Important
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Ex 13.2, 9 (iv) Important
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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
Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)
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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
