Example 6 - Find derivative of f(x) = 2x2 + 3x - 5  at x =  -1 - Examples

Example 6  - Chapter 13 Class 11 Limits and Derivatives - Part 2
Example 6  - Chapter 13 Class 11 Limits and Derivatives - Part 3
Example 6  - Chapter 13 Class 11 Limits and Derivatives - Part 4
Example 6  - Chapter 13 Class 11 Limits and Derivatives - Part 5

Example 6  - Chapter 13 Class 11 Limits and Derivatives - Part 6 Example 6  - Chapter 13 Class 11 Limits and Derivatives - Part 7

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Example ,6 (Method 1) Find the derivative of the function f(x) = 2x2 + 3x – 5 at x = –1. Also prove that f’(0) + 3f’( –1) = 0. Given f(x) = 2x2 + 3x – 5 We know that f’(x) = lim﷮h→0﷯ f﷮ 𝑥 + ℎ﷯ − f (x)﷯﷮h﷯ Now f (x) = 2x2 + 3x – 5 So, f (x + h) = 2(x + h)2 + 3(x + h) – 5 Putting values f’ (x) = lim﷮h→0﷯ (2(𝑥 + ℎ)﷮2﷯ + 3 𝑥 + ℎ﷯ − 5) − (2𝑥﷮2﷯ + 3𝑥 − 5)﷮h﷯ f’ (x) = lim﷮h→0﷯ (2(𝑥 + ℎ)﷮2﷯ + 3 𝑥 + ℎ﷯ − 5) − (2𝑥﷮2﷯ + 3𝑥 − 5)﷮h﷯ Putting x = – 1 f’( –1) = lim﷮h→0﷯ (2(−1+ℎ)﷮2﷯ + 3 −1+ℎ﷯ − 5) − (2 ((−1)﷮2﷯) + 3(−1) − 5)﷮h﷯ = lim﷮h→0﷯ (2(−1+ℎ)﷮2﷯ + 3 −1+ℎ﷯ − 5) − (2 1﷯ − 3 − 5)﷮h﷯ = lim﷮h→0﷯ (2(−1+ℎ)﷮2﷯ + 3 −1+ℎ﷯ − 5) − (−6)﷮h﷯ = lim﷮h→0﷯ 2(−1+ℎ)﷮2﷯ + 3 −1+ℎ﷯ − 5 + 6﷮h﷯ = lim﷮h→0﷯ 2 −1﷯2 + ℎ2 +2 −1﷯ℎ﷯ − 3 + 3ℎ + 1﷮h﷯ = lim﷮h→0﷯ 2 1 + ℎ2 − 2ℎ﷯ + 3ℎ − 2﷮h﷯ = lim﷮h→0﷯ 2 + 2ℎ2 − 4ℎ + 3ℎ − 2﷮h﷯ = lim﷮h→0﷯ 2ℎ2− ℎ﷮h﷯ = lim﷮h→0﷯ ℎ(2ℎ − 1)﷮h﷯ = lim﷮h→0﷯ 2h – 1 Putting h = 0 = 2(0) – 1 = – 1 Hence f’( –1) = – 1 Now, finding f’(0) For f’(0) f’(x)= lim﷮h→0﷯ 𝑓 𝑥 + ℎ﷯ − 𝑓(𝑥)﷮ℎ﷯ f’(x)= lim﷮h→0﷯ 2 𝑥 + ℎ﷯2 + 3 𝑥 + ℎ﷯ − 5﷯ −[2𝑥2 + 3𝑥 − 5]﷮ℎ﷯ putting x = 0 f’(0)= lim﷮h→0﷯ 2 0 + ℎ﷯2 + 3 0 + ℎ﷯ − 5﷯ −[2(0)2 + 3(0) − 5]﷮ℎ﷯ f’(0)= lim﷮h→0﷯ 2ℎ2 + 3ℎ − 5﷯ − [0 + 0 − 5]﷮ℎ﷯ f’(0)= lim﷮h→0﷯ 2ℎ2 + 3ℎ − 5 + 5﷮ℎ﷯ f’(0)= lim﷮h→0﷯ 2ℎ2 + 3ℎ﷮ℎ﷯ = lim﷮h→0﷯ ℎ(2ℎ+3)﷮h﷯ = lim﷮h→0﷯ 2h + 3 Putting h = 0 = 2(0) + 3 = 3 Hence, f’(0) = 3 Now, f’ (0) + 3f’( – 1) Putting value of f’(0) & f’( –1) = 3 + 3 ( –1) = 0 Hence Proved Example 6 (Method 2) Find the derivative of the function f(x) = 2x2 + 3x – 5 at x = –1. Also prove that f’(0) + 3f’( –1) = 0. Given f(x) = 2x2 + 3x – 5 Now, f’(x) = (2x2 + 3x – 5)’ = 2(2.x2–1) + 3(1.x1–1) – 0 = 2(2x1) + 3(1) = 4x + 3 Putting x = 0 f’(0) = 4(0) + 3 = 0 + 3 = 3 Taking f’ (0) + 3f’( – 1) Putting value of f’(0) & f’( –1) = 3 + 3 ( –1) = 3 – 3 = 0 Hence Proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo