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

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 Ex 13.2, 4 Find the derivative of the following functions from first principle. (ii) (x – 1) (x – 2) Let f (x) = (x – 1) (x – 2) = x(x – 2) – 1 (x – 2) = x2 – 2x – x + 2 = x2 – 3x + 2 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﷯ Here, f(x) = x2 – 3x + 2 So, f(x + h) = (x + h)2 – 3 (x + h) + 2 Putting values f’ (x) = lim﷮h→0﷯﷮ x + h﷯2 − 3 x + h﷯ + 2﷯ − (x2 − 3 + 2)﷮h﷯﷯ = lim﷮h→0﷯﷮ x + h﷯2 − 3x − 3h + 2 − x2 + 3x − 2﷮h﷯﷯ = lim﷮h→0﷯﷮ x + h﷯2 −3h − x2 + 3𝑥 − 3𝑥 + 2 − 2﷮h﷯﷯ = lim﷮h→0﷯﷮ x + h﷯2 − x2 −3h ﷮h﷯﷯ = lim﷮h→0﷯﷮ x2 + h2+ 2xh − x2 −3h ﷮h﷯﷯ = lim﷮h→0﷯﷮ ℎ2 + 2𝑥ℎ − 3ℎ − 𝑥2 + 𝑥2﷮h﷯﷯ = lim﷮h→0﷯﷮ h h + 2𝑥 − 3﷯ ﷮h﷯﷯ = lim﷮h→0﷯﷮ℎ+2𝑥 −3﷯ Putting h = 0 = 0 + 2x – 3 = 2x – 3 Hence f’ (x) = 2x – 3 Ex 13.2, 4 Find the derivative of the following functions from first principle. (iii) 1﷮ 𝑥﷮2﷯﷯ Let f (x) = 1﷮ 𝑥﷮2﷯﷯ 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﷯ Here, f (x) = 1﷮ 𝑥﷮2﷯﷯ So, f (x + h) = 1﷮ (𝑥 + ℎ)﷮2﷯﷯ Putting values f’(x) = lim﷮h→0﷯﷮ 1﷮ (𝑥 + ℎ)﷮2﷯﷯ − 1﷮ 𝑥﷮2﷯﷯﷮h﷯﷯ = lim﷮h→0﷯﷮ 𝑥 ﷮2﷯ − (𝑥 + ℎ)﷮2﷯﷮ (𝑥 + ℎ)﷮2﷯ 𝑥﷮2﷯﷯﷮h﷯﷯ = lim﷮h→0﷯﷮ x2 − x + h﷯2﷮hx2 x + h﷯2﷯﷯ = lim﷮h→0﷯﷮ 𝑥 − 𝑥 + ℎ ﷯﷯ 𝑥 + 𝑥 + ℎ﷯﷯﷮hx2 x + h﷯2﷯﷯ = lim﷮h→0﷯﷮ x −x − h﷯ (𝑥 + 𝑥 + ℎ)﷮h.x2 x + h﷯2﷯﷯ = lim﷮h→0﷯﷮ − ℎ﷯(2𝑥 + ℎ)﷮hx2 x + h﷯2﷯﷯ = lim﷮h→0﷯﷮ −1﷯ 2x + h﷮x2 x + h﷯2﷯﷯ Putting h = 0 = −1﷯ 2𝑥 + 0﷮𝑥2 𝑥 + 0﷯2﷯ = −1﷯ 2𝑥﷮𝑥2 𝑥﷯2﷯ = −2𝑥﷮𝑥4﷯ = −2﷮𝑥3﷯ Thus, f’(x) = −𝟐﷮𝒙𝟑﷯ Ex 13.2, 4 Find the derivative of the following functions from first principle. (iv) 𝑥 + 1﷮𝑥 − 1﷯ Let f (x) = 𝑥 + 1﷮𝑥 − 1﷯ 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﷯ Here, f (x) = 𝑥 + 1﷮𝑥 − 1﷯ So, f (x + h) = 𝑥 + ℎ﷯ + 1﷮ 𝑥 + ℎ﷯ − 1﷯ Putting values f’(x) = lim﷮h→0﷯﷮ 𝑥 + ℎ﷯ + 1﷮𝑥 + ℎ − 1﷯﷯ − 𝑥 + 1﷮𝑥 − 1﷯﷯﷮h﷯﷯ = lim﷮h→0﷯﷮ 𝑥 + ℎ + 1﷮𝑥 + ℎ − 1﷯ − 𝑥 + 1﷮𝑥 − 1﷯﷮h﷯﷯ = lim﷮h→0﷯﷮ 𝑥 − 1﷯ 𝑥 + ℎ + 1﷯ − 𝑥 + 1﷯( 𝑥 + ℎ − 1)﷮ 𝑥 + ℎ − 1 ﷯ (𝑥 − 1)﷯﷮ℎ﷯﷯ = lim﷮h→0﷯﷮ 𝑥 − 1﷯ 𝑥 + 1﷯ + ℎ﷯ − 𝑥 + 1﷯( 𝑥 − 1﷯ + ℎ)﷮ℎ 𝑥 + ℎ − 1 ﷯ (𝑥 − 1)﷯﷯ = lim﷮h→0﷯﷮ 𝑥 − 1﷯ 𝑥 + 1﷯ + 𝑥 −1﷯ℎ− 𝑥 + 1﷯ 𝑥 − 1﷯ −(𝑥 + 1) ℎ﷮ℎ 𝑥 + ℎ − 1 ﷯ (𝑥 − 1)﷯﷯ = lim﷮h→0﷯﷮ 𝑥2 − 1﷯ + 𝑥ℎ − ℎ 𝑥2 − 1﷯ − 𝑥ℎ − ℎ﷮ℎ 𝑥 + ℎ − 1 ﷯ (𝑥 − 1)﷯﷯ = lim﷮h→0﷯﷮ − ℎ − ℎ + 𝑥2 − 𝑥2 − 1 + 1 + 𝑥ℎ − 𝑥ℎ﷮ℎ 𝑥 + ℎ − 1﷯ (𝑥 − 1)﷯﷯ = lim﷮h→0﷯﷮ − 2h + 0 + 0 + 0﷮ℎ 𝑥 + ℎ − 1﷯ (𝑥 − 1)﷯﷯ = lim﷮h→0﷯﷮ − 2h ﷮ℎ 𝑥 + ℎ − 1﷯ (𝑥 − 1)﷯﷯ = lim﷮h→0﷯﷮ −2﷮ 𝑥 + ℎ − 1﷯ (𝑥 − 1)﷯﷯ Putting h = 0 = −2﷮ 𝑥 + 0 − 1﷯(𝑥 − 1)﷯ = −2﷮ 𝑥 − 1﷯ (𝑥 − 1)﷯ = −2﷮ 𝑥 − 1﷯2﷯ Hence, f’(x) = −𝟐﷮ 𝒙 − 𝟏﷯𝟐﷯