Misc 1 - Differentiate (3x2 - 9x + 5)9 - Chapter 5 Class 12 - Finding derivative of a function by chain rule

Misc 1 last slide.jpg

  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Concept wise

Transcript

Misc 1 Differentiate w.r.t. x the function, (3๐‘ฅ2 โ€“ 9๐‘ฅ + 5)^9 Let ๐‘ฆ=(3๐‘ฅ2 โ€“ 9๐‘ฅ + 5)^9 Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘(3๐‘ฅ2 โ€“ 9๐‘ฅ + 5)^9)/๐‘‘๐‘ฅ = 9(3๐‘ฅ2 โ€“ 9๐‘ฅ + 5)^(9 โˆ’1) . ๐‘‘(3๐‘ฅ2 โ€“ 9๐‘ฅ + 5)/๐‘‘๐‘ฅ = 9(3๐‘ฅ2 โ€“ 9๐‘ฅ + 5)^8 . (๐‘‘(3๐‘ฅ2)/๐‘‘๐‘ฅ โˆ’ ๐‘‘(9๐‘ฅ)/๐‘‘๐‘ฅ โˆ’ ๐‘‘(5)/๐‘‘๐‘ฅ) = 9(3๐‘ฅ2 โ€“ 9๐‘ฅ + 5)^8 . (6๐‘ฅโˆ’9+0) = 9(3๐‘ฅ2 โ€“ 9๐‘ฅ + 5)^8 . (6๐‘ฅโˆ’9) = 9(3๐‘ฅ2 โ€“ 9๐‘ฅ + 5)^8 . 3(2๐‘ฅโˆ’3) = 27(3๐‘ฅ2 โ€“ 9๐‘ฅ + 5)^8 (2๐‘ฅโˆ’3) Hence, ๐’…๐’š/๐’…๐’™ = ๐Ÿ๐Ÿ•(๐Ÿ‘๐’™๐Ÿ โ€“ ๐Ÿ—๐’™ + ๐Ÿ“)^๐Ÿ– (๐Ÿ๐’™โˆ’๐Ÿ‘)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.