Ex 13.2, 9 - Class 11

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Ex 13.2, 9 Find the derivative of (i) 2𝑥 – 3﷮4﷯ Let f(x) = 2𝑥 – 3﷮4﷯ f’(x) = 𝑑(2𝑥 − 3﷮4﷯ )﷮𝑑𝑥﷯ = 2 – 0 = 2 ∴ f’(x) = 2 Ex 13.2, 9 Find the derivative of (ii) (5x3 + 3x – 1) (x – 1) Let f(x) = (5x3 + 3x – 1) (x- 1) Let u = 5x3 + 3x – 1 & v = x – 1 ∴ f(x) = uv So, f’(x) = (uv)’ f’(x) = u’v + v’u Finding u’ & v’ u = 5x3 + 3x – 1 u’ = 5.3x3–1 + 3.1.x1–1 – 0 = 15x2 + 3x0 = 15x2 + 3 v = x – 1 v’ = 1 .x1–1 – 0 = 1. x0 = 1 f’(x) = u’v + v’u = (15x2 + 3) (x – 1) + (1) (5x3 + 3x – 1) = 15x2 (x – 1) + 3 (x – 1) + 5x3 + 3x – 1 = 15x3 – 15x2 + 3x – 3 + 5x3 + 3x – 1 = 15x3 + 5x3 – 15x2 + 3x + 3x – 3 – 1 = 20x3 – 15x2 + 6x – 4 Hence f’(x) = 20x3 – 15x2 + 6x – 4 Ex 13.2, 9 Find the derivative of (iii) x–3 (5 + 3x) Let f(x) = x – 3 (5 + 3x) Let u = x – 3 & v = 5 + 3x ∴ f(x) = uv So, f’(x) = (uv)’ f’(x) = u’v + v’u Finding u’ & v’ u = x – 3 u’ = – 3x – 3 – 1 = – 3x – 4 v = 5 + 3x v’ = 0 + 3 = 3 Now, f’(x) = (uv)’ = u’v + v’u = – 3x – 4 (5 + 3x) + 3 (x – 3) = – 15x – 4 – 9x – 4+ 1 + 3x – 3 = – 15x – 4 – 6x – 3 = −15﷮ 𝑥﷮4﷯﷯ – 6﷮ 𝑥﷮3﷯﷯ = −3﷮ 𝑥﷮4﷯﷯ (5 + 2x) Hence f’(x) = −𝟑﷮ 𝒙﷮𝟒﷯﷯ (5 + 2x) Ex 13.2,9 Find the derivative of (iv) x5 (3 − 6x−9 ). Let f (x) = x5 (3 − 6x−9 ) Let u = x5 & v = 3 – 6x – 9 So, f(x) = uv ∴ f’ (x) = (uv)’ f’ (x) = u’v + v’ u Finding u’ & v’ u = x5 u’ = 5x5 – 1 = 5x4 v = 3 − 6x−9 v’ = 0 – 6( – 9)x – 9 – 1 v’ = 54x –10 Now, f’(x) = (uv)’ = u’v + v’ u = 5x4 (3 – 6x – 9) + 54x – 10) (x5) = 15x4 – 30x – 9 + 4 + 54 – 10 + 5 = 15x4 – 30x –5 + 54x – 5 = 15x4 + 24x – 5 = 15x4 + 24﷮ 𝑥﷮5﷯﷯ Hence f’(x) = 15x4 + 𝟐𝟒﷮ 𝒙﷮𝟓﷯﷯ Ex 13.2, 9 Find the derivative of (v) f (x) = x–4 (3 – 4x–5) Let f(x) = x – 4 (3 – 4x – 5) Let u = x – 4 & v = 3 – 4x – 5 ∴ f(x) = uv So, f’(x) = (uv)’ f’(x) = u’v + v’u Finding u’ & v’ u = x – 4 u’ = – 4x – 4 – 1 = – 4x – 5 v = 3 – 4x – 5 v’ = 0 – 4 ( – 5) x – 5 – 1 = 20x – 6 Now, f’(x) = (uv)’ = u’v + v’u = (– 4x–5 ) (3 – 4x–5) + (20x–6) (x–4) = – 12x–5 + 16x –5 – 5 + 20 –6 – 4 = – 12x –5 + 16x –10 + 20x – 10 = – 12x –5 + 36x – 10 = −12﷮ 𝑥﷮5﷯﷯ + 36﷮ 𝑥﷮10﷯﷯ Hence f’(x) = −𝟏𝟐﷮ 𝒙﷮𝟓﷯﷯ + 𝟑𝟔﷮ 𝒙﷮𝟏𝟎﷯﷯ 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 ∴ f2 (x) = 𝑢﷮𝑣﷯ 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﷯﷯﷮′﷯ = 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)﷯