Last updated at Aug. 28, 2021 by

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Ex 13.2, 9 Find the derivative of (vi) f(x) = 2/(x + 1) – x2/(3x − 1) Let f (x) = 2/(x + 1) – x2/(3x − 1) Let f1 (x) = 2/(x + 1) & f2 (x) = x2/(3x − 1) ∴ f(x) = f1(x) – f2 (x) So, f’(x) = (f1(x) – f2(x))’ f’(x) = f’1(x) – f’2(x) Finding f1‘(x) f1 (x) = 2/(𝑥 + 1) Let u = 2 & v = x + 1 ∴ f1(x) = 𝑢/𝑣 Now, f1’(x) = (𝑢/𝑣)^′ f1’(x) = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 u = 2 u’ = 0 v = x + 1 v’ = 1 + 0 = 1 f’1(x) = (𝑢/𝑣)^′ = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 = (0(𝑥 + 1) −1 (2))/(𝑥 + 1)2 = (−2)/〖(𝑥 + 1)〗^2 Hence, f1’ (x) = (−2)/(𝑥 + 1)2 Finding f2‘(x) f2 (x) = 𝑥2/(3𝑥 − 1) Let u = x2 & v = 3x – 1 Now, f2’(x) = (𝑢/𝑣)^′ f2’(x) = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 Finding u’ & v’ u = x2 u’ = 2x2 – 1 = 2x & v = 3x – 1 v’ = 3(1) – 0 = 3 f’2(x) = (𝑢/𝑣)^′ (xn)’ = nxn – 1 & (a)’ = 0 where a is constant = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 = (2𝑥(3𝑥 − 1) − 3 (𝑥2))/(3𝑥 − 1)2 = (6𝑥2 − 2𝑥 − 3𝑥2)/〖(3𝑥 − 1)〗^2 = (3𝑥2 − 2𝑥 )/〖(3𝑥 − 1)〗^2 = (𝑥(3𝑥 − 2))/〖(3𝑥 − 1)〗^2 Hence f’2(x) = (𝑥 (3𝑥 − 2))/(3𝑥 − 1)2 Now f’ (x) = f1’(x) – f2’ (x) = (−𝟐)/(𝒙 + 𝟏)𝟐 – (𝒙(𝟑𝒙 − 𝟐))/(𝟑𝒙 − 𝟏)𝟐

Ex 13.2 (Term 2)

Ex 13.2, 1

Ex 13.2, 2

Ex 13.2, 3

Ex 13.2, 4 (i) Important

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) You are here

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

Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.