Example 44 - Differentiate (i) root 3x + 2 - Chapter 5 Class 12 - Examples

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  1. Class 12
  2. Important Question for exams Class 12

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) = 3/(2โˆš(3๐‘ฅ + 2 )) Calculating (๐’…(๐Ÿ๐’™^๐Ÿ + ๐Ÿ’)^((โˆ’๐Ÿ)/๐Ÿ))/๐’…๐’™ (๐‘‘(2๐‘ฅ + 4)^((โˆ’1)/2))/๐‘‘๐‘ฅ = (โˆ’1)/2 ใ€–(2๐‘ฅ+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+ 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’2๐‘ฅ)/(2๐‘ฅ^2+ 4)^(3/2) 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)) Example 44 Differentiate w.r.t. x, the following function: (ii) ๐‘’^(sec^2โก๐‘ฅ ) + 3cos^(โ€“1) ๐‘ฅ Let y = ๐‘’^(sec^2โก๐‘ฅ ) + 3cos^(โ€“1) ๐‘ฅ Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(๐‘’^(sec^2โก๐‘ฅ )+ 3cos^(โ€“1) ๐‘ฅ" " )/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(๐‘’^(sec^2โก๐‘ฅ ) )/๐‘‘๐‘ฅ + ๐’…(๐Ÿ‘ใ€–๐’„๐’๐’”ใ€—^(โ€“๐Ÿ) ๐’™)/๐’…๐’™ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘’^(sec^2โก๐‘ฅ ) ๐‘‘(sec^2โก๐‘ฅ )/๐‘‘๐‘ฅ + 3. ((โˆ’๐Ÿ)/โˆš(๐Ÿ โˆ’๐’™^๐Ÿ )) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘’^(sec^2โก๐‘ฅ ). 2 sec ๐‘ฅ . ๐’…(ใ€–๐’”๐’†๐’„ ใ€—โก๐’™ )/๐’…๐’™ โˆ’ 3/โˆš(1 โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘’^(sec^2โก๐‘ฅ ). 2 sec ๐‘ฅ . ๐’”๐’†๐’„โก๐’™ .๐’•๐’‚๐’โก๐’™ โˆ’ 3/โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐’…๐’š/๐’…๐’™ = ใ€–๐Ÿ ๐’†ใ€—^(ใ€–๐’”๐’†๐’„ใ€—^๐Ÿโก๐’™ ). ใ€–๐’”๐’†๐’„ใ€—^๐Ÿ ๐’™ .๐’•๐’‚๐’โก๐’™ โˆ’ ๐Ÿ‘/โˆš(๐Ÿ โˆ’ ๐’™^๐Ÿ ) Example 44 Differentiate w.r.t. x, the following function: (iii) log7 (log x) y = log7 (log x) But we do not solve base 7. So, changing base y = log7 ( log x) y = ๐ฅ๐จ๐ โกใ€–(๐ฅ๐จ๐ โกใ€–๐’™)ใ€— ใ€—/๐ฅ๐จ๐ โก๐Ÿ• Now, differentiating ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= ๐‘‘(((logโกใ€–(logโกใ€–7)ใ€— ใ€—)/(log 7))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= ๐Ÿ/๐ฅ๐จ๐ โก๐Ÿ• (๐‘‘(logโกใ€–(logโก๐‘ฅ ใ€— )) )/๐‘‘๐‘ฅ y = log7 (log x) But we do not solve base 7. So, changing base y = log7 ( log x) y = ๐ฅ๐จ๐ โกใ€–(๐ฅ๐จ๐ โกใ€–๐’™)ใ€— ใ€—/๐ฅ๐จ๐ โก๐Ÿ• Now, differentiating ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= ๐‘‘(((logโกใ€–(logโกใ€–7)ใ€— ใ€—)/(log 7))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= ๐Ÿ/๐ฅ๐จ๐ โก๐Ÿ• (๐‘‘(logโกใ€–(logโก๐‘ฅ ใ€— )) )/๐‘‘๐‘ฅ

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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.