# Ex 13.2, 9 (iv) - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

Last updated at Sept. 6, 2021 by Teachoo

Last updated at Sept. 6, 2021 by Teachoo

Transcript

Ex 13.2,9 (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 13.2, 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

Ex 13.2 (Term 2)

Ex 13.2, 1

Ex 13.2, 2

Ex 13.2, 3

Ex 13.2, 4 (i) Important

Ex 13.2, 4 (ii)

Ex 13.2, 4 (iii) Important

Ex 13.2, 4 (iv)

Ex 13.2, 5

Ex 13.2, 6

Ex 13.2, 7 (i) Important

Ex 13.2, 7 (ii)

Ex 13.2, 7 (iii) Important

Ex 13.2, 8

Ex 13.2, 9 (i)

Ex 13.2, 9 (ii) Important

Ex 13.2, 9 (iii)

Ex 13.2, 9 (iv) Important You are here

Ex 13.2, 9 (v)

Ex 13.2, 9 (vi)

Ex 13.2, 10 Important

Ex 13.2, 11 (i)

Ex 13.2, 11 (ii) Important

Ex 13.2, 11 (iii) Important

Ex 13.2, 11 (iv)

Ex 13.2, 11 (v) Important

Ex 13.2, 11 (vi)

Ex 13.2, 11 (vii) Important

Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.