Ex 12.2, 9 (iv) - Chapter 12 Class 11 Limits and Derivatives

Last updated at Dec. 16, 2024 by

Ex 13.2, 9 - Chapter 13 Class 11 Limits and Derivatives - Part 10

Ex 13.2, 9 - Chapter 13 Class 11 Limits and Derivatives - Part 11
Ex 13.2, 9 - Chapter 13 Class 11 Limits and Derivatives - Part 12
Ex 13.2, 9 - Chapter 13 Class 11 Limits and Derivatives - Part 13

Remove Ads

Transcript

Ex 12.2,9 (Method 1) 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 u’ = 5x4 v = 3 − 6x−9 v’ = 0 – 6( –9)x–10 v’ = 54x–10 Now, f’(x) = (uv)’ = u’v + v’ u = 5x4 (3 – 6x–9) + 54x–10 (x5) = 15x4 – 30x–9 + 4 + 54x–10 + 5 = 15x4 – 30x –5 + 54x –5 = 15x4 + 24x –5 = 15x4 + 24x –5 = 15x4 + 24/𝑥^5 Hence f’(x) = 15x4 + 𝟐𝟒/𝒙^𝟓 Ex 12.2, 9 (Method 2) Find the derivative of (iv) x5 (3 − 6x−9 ). x5 (3 − 6x−9 ). = 〖3𝑥〗^5 − 〖6𝑥〗^(−4) Differentiating w.r.t.x 〖15𝑥〗^4 −6 [−4𝑥^(−5)] =〖15𝑥〗^4+24𝑥^(−5) = 15x4 + 𝟐𝟒/𝒙^𝟓

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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo