Example 44 - Differentiate the function (i) root (3x+2) + 1/root(2x^2

Example 44 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Example 44 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Example 44 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

 

 

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Example 44 Differentiate w.r.t. x, the following function: (i) โˆš(3๐‘ฅ+2) + 1/โˆš(2๐‘ฅ^2+ 4) Let y = โˆš(3๐‘ฅ+2) + 1/โˆš(2๐‘ฅ^2+ 4 ) Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(โˆš(3๐‘ฅ + 2) " + " 1/โˆš(2๐‘ฅ^2 + 4 ))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(โˆš(3๐‘ฅ + 2))/๐‘‘๐‘ฅ + ๐‘‘(1/โˆš(2๐‘ฅ^2 + 4 ))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(โˆš(3๐‘ฅ + 2))/๐‘‘๐‘ฅ + (๐‘‘(2๐‘ฅ^2 + 4)^((โˆ’1)/2))/๐‘‘๐‘ฅ Calculating ๐‘‘(โˆš(3๐‘ฅ + 2))/๐‘‘๐‘ฅ & (๐‘‘(2๐‘ฅ^2 + 4)^((โˆ’1)/2))/๐‘‘๐‘ฅ separately Calculating ๐(โˆš(๐Ÿ‘๐ฑ + ๐Ÿ))/๐’…๐’™ ๐‘‘(โˆš(3๐‘ฅ + 2))/๐‘‘๐‘ฅ = 1/(2โˆš(3๐‘ฅ + 2 )) ร— ๐‘‘(3๐‘ฅ + 2)/๐‘‘๐‘ฅ = 1/(2โˆš(3๐‘ฅ + 2 )) ร— (3+0) = ๐Ÿ‘/(๐Ÿโˆš(๐Ÿ‘๐’™ + ๐Ÿ )) Calculating (๐’…(๐Ÿ๐’™^๐Ÿ + ๐Ÿ’)^((โˆ’๐Ÿ)/๐Ÿ))/๐’…๐’™ (๐‘‘(2๐‘ฅ^2 + 4)^((โˆ’1)/2))/๐‘‘๐‘ฅ = (โˆ’1)/2 ใ€–(2๐‘ฅ^2+4)ใ€—^((โˆ’1)/( 2) โˆ’1) . ๐‘‘(2๐‘ฅ^2+ 4)/๐‘‘๐‘ฅ = (โˆ’1)/2 (2๐‘ฅ^2+ 4)^((โˆ’3)/( 2)) . (๐‘‘(2๐‘ฅ^2 )/๐‘‘๐‘ฅ + ๐‘‘(4)/๐‘‘๐‘ฅ) = (โˆ’1)/2 (2๐‘ฅ^2+ 4)^((โˆ’3)/( 2)) . (4๐‘ฅ+0) = (โˆ’4๐‘ฅ)/2 (2๐‘ฅ^2+ 4)^((โˆ’3)/( 2)) = (โˆ’๐Ÿ๐’™)/(๐Ÿ๐’™^๐Ÿ+ ๐Ÿ’)^(๐Ÿ‘/๐Ÿ) Hence, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(โˆš(3๐‘ฅ+2))/๐‘‘๐‘ฅ + ๐‘‘(1/โˆš(2๐‘ฅ^2+ 4 ))/๐‘‘๐‘ฅ ๐’…๐’š/๐’…๐’™ = ๐Ÿ‘/(๐Ÿโˆš(๐Ÿ‘๐’™ + ๐Ÿ )) โˆ’ ๐Ÿ๐’™/(๐Ÿ๐’™^๐Ÿ+ ๐Ÿ’)^(๐Ÿ‘/( 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 10 years. He provides courses for Maths and Science at Teachoo.