Check sibling questions

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


Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.