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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
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
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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.