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Ex 5.5, 11 - Differentiate (x cos x)x + (x sin x)1/x - Ex 5.5

Ex 5.5, 11 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5, 11 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.5, 11 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.5, 11 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Ex 5.5, 11 - Chapter 5 Class 12 Continuity and Differentiability - Part 6
Ex 5.5, 11 - Chapter 5 Class 12 Continuity and Differentiability - Part 7

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Transcript

Ex 5.5, 11 Differentiate the functions in, γ€–(π‘₯ π‘π‘œπ‘ β‘π‘₯ ) γ€—^π‘₯ + γ€–(π‘₯ 𝑠𝑖𝑛⁑π‘₯ ) γ€—^(1/π‘₯)𝑦 = γ€–(π‘₯ π‘π‘œπ‘ β‘π‘₯ ) γ€—^π‘₯ + γ€–(π‘₯ 𝑠𝑖𝑛⁑π‘₯ ) γ€—^(1/π‘₯) Let 𝑒 = γ€–(π‘₯ π‘π‘œπ‘ β‘π‘₯ ) γ€—^π‘₯ , 𝑣 = γ€–(π‘₯ 𝑠𝑖𝑛⁑π‘₯ ) γ€—^(1/π‘₯) 𝑦 = 𝑒+𝑣 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑π‘₯ = (𝑑 (𝑒 + 𝑣))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Calculating 𝒅𝒖/𝒅𝒙 𝑒 =γ€– (π‘₯ π‘π‘œπ‘ β‘π‘₯ ) γ€—^π‘₯ Taking log both sides . log⁑𝑒 = logγ€– (π‘₯ π‘π‘œπ‘ β‘π‘₯ ) γ€—^π‘₯ log⁑𝑒 = π‘₯ . log (π‘₯ π‘π‘œπ‘ β‘π‘₯ ) Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑒 )/𝑑π‘₯ = (𝑑(π‘₯ . log⁑(π‘₯ cos⁑π‘₯ ) ) )/𝑑π‘₯ 𝑑(log⁑𝑒 )/𝑑𝑒 . 𝑑𝑒/𝑑π‘₯ = (𝑑(π‘₯ . log⁑(π‘₯ cos⁑π‘₯ ) ) )/𝑑π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯) = (𝑑(π‘₯ . log⁑(π‘₯ cos⁑π‘₯ ) ) )/𝑑π‘₯ (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏 )=𝑏 . π‘™π‘œπ‘”β‘π‘Ž) Using product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 1/𝑒 (𝑑𝑒/𝑑π‘₯) = 𝑑π‘₯/𝑑π‘₯ log⁑〖(π‘₯ π‘π‘œπ‘  π‘₯)γ€—+π‘₯ (𝑑(π‘™π‘œπ‘”β‘(π‘₯ π‘π‘œπ‘ β‘π‘₯ ) ) )/𝑑π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯) = log⁑〖(π‘₯ cos⁑π‘₯)γ€—+π‘₯/(π‘₯ cos⁑π‘₯ ) Γ— (π‘₯ π‘π‘œπ‘  π‘₯)^β€² 1/𝑒 (𝑑𝑒/𝑑π‘₯) = log⁑〖(π‘₯ cos⁑π‘₯)γ€—+1/cos⁑π‘₯ Γ— (1.cos⁑π‘₯+π‘₯(βˆ’sin⁑π‘₯ )) 1/𝑒 (𝑑𝑒/𝑑π‘₯) = log⁑〖(π‘₯ cos⁑π‘₯)γ€—+((cos⁑π‘₯ βˆ’ π‘₯ sin⁑π‘₯ ))/cos⁑π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯) = log⁑〖(π‘₯ cos⁑π‘₯)γ€—+cos⁑π‘₯/cos⁑π‘₯ βˆ’(π‘₯ sin⁑π‘₯)/cos⁑π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯) = log⁑〖(π‘₯ cos⁑π‘₯)γ€—+1βˆ’π‘₯ π‘‘π‘Žπ‘› π‘₯ 𝑑𝑒/𝑑π‘₯ = u (1βˆ’π‘₯ tan⁑π‘₯+log⁑〖(π‘₯ cos⁑π‘₯ γ€—)) Putting value of 𝑒 𝑑𝑒/𝑑π‘₯ = (π‘₯ cos⁑π‘₯ )^π‘₯ (1βˆ’π‘₯ tan⁑π‘₯+π’π’π’ˆβ‘(𝒙 𝒄𝒐𝒔⁑𝒙 ) ) Calculating 𝒅𝒗/𝒅𝒙 𝑣=γ€–(π‘₯ 𝑠𝑖𝑛⁑π‘₯ ) γ€—^(1/π‘₯) Taking log both sides log⁑𝑣=log⁑〖 γ€–(π‘₯ 𝑠𝑖𝑛⁑π‘₯ ) γ€—^(1/π‘₯) γ€— log⁑𝑣= 1/π‘₯ log (π‘₯ 𝑠𝑖𝑛⁑π‘₯ ) Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. (𝑑(log⁑𝑣))/𝑑π‘₯ = 𝑑(1/π‘₯ " . " )/𝑑π‘₯ (𝑑(log⁑𝑣))/𝑑𝑣 . 𝑑𝑣/𝑑π‘₯ = (1/π‘₯ .log⁑(π‘₯ sin⁑π‘₯ ) )^β€² 1/𝑣 Γ— 𝑑𝑣/𝑑π‘₯ = (1/π‘₯ .log⁑(π‘₯ sin⁑π‘₯ ) )^β€² 1/𝑣 Γ— 𝑑𝑣/𝑑π‘₯ = (((log⁑〖(π‘₯ sin⁑π‘₯)γ€— )^β€² π‘₯ + π‘₯^β€² log⁑〖(π‘₯ sin⁑π‘₯)γ€—)/π‘₯^2 ) Using quotient rule (𝑒/𝑣)^β€² = (𝑒^β€² 𝑣 βˆ’ 𝑣^β€² 𝑒)/𝑣^2 Where u = log (x sin x) , v = x 1/𝑣 Γ— 𝑑𝑣/𝑑π‘₯ = 1/x^2 ((log⁑〖(π‘₯ sin⁑π‘₯)γ€— )^β€² π‘₯ + log⁑〖(π‘₯ sin⁑π‘₯)γ€— ) 1/𝑣 Γ— 𝑑𝑣/𝑑π‘₯ = 1/x^2 ([1/(π‘₯ sin⁑π‘₯ ) \ Γ—(π‘₯ sin⁑π‘₯ )^β€² ]π‘₯ + log⁑〖(π‘₯ sin⁑π‘₯)γ€— ) 1/𝑣 Γ— 𝑑𝑣/𝑑π‘₯ = 1/x^2 (1/sin⁑π‘₯ \ Γ—(π‘₯ sin⁑π‘₯ )^β€²+ log⁑〖(π‘₯ sin⁑π‘₯)γ€— ) 1/𝑣 Γ— 𝑑𝑣/𝑑π‘₯ = 1/π‘₯^2 (((1 . 𝑠𝑖𝑛⁑π‘₯ + π‘₯ π‘π‘œπ‘ β‘π‘₯))/𝑠𝑖𝑛⁑π‘₯ \ + π‘™π‘œπ‘”β‘γ€–(π‘₯ 𝑠𝑖𝑛⁑π‘₯)γ€— ) 1/𝑣 Γ— 𝑑𝑣/𝑑π‘₯ = 1/π‘₯^2 (((𝑠𝑖𝑛⁑π‘₯ + π‘₯ π‘π‘œπ‘ β‘π‘₯))/𝑠𝑖𝑛⁑π‘₯ \ + π‘™π‘œπ‘”β‘γ€–(π‘₯ 𝑠𝑖𝑛⁑π‘₯)γ€— ) 1/𝑣 Γ— 𝑑𝑣/𝑑π‘₯ = 1/π‘₯^2 (1+π‘₯ cot⁑π‘₯+ π‘™π‘œπ‘”β‘γ€–(π‘₯ 𝑠𝑖𝑛⁑π‘₯)γ€— ) 𝑑𝑣/𝑑π‘₯ = 𝑣((1 + γ€–π‘₯ π‘π‘œπ‘‘γ€—β‘γ€–π‘₯ γ€— βˆ’ π‘™π‘œπ‘”β‘γ€– (π‘₯ 𝑠𝑖𝑛⁑π‘₯ ) γ€—)/π‘₯^2 ) 𝑑𝑣/𝑑π‘₯ = (π‘₯ 𝑠𝑖𝑛⁑π‘₯ )^(1/π‘₯) ((1 + γ€–π‘₯ π‘π‘œπ‘‘γ€—β‘γ€–π‘₯ γ€—βˆ’ π‘™π‘œπ‘” (π‘₯ 𝑠𝑖𝑛⁑π‘₯ ))/π‘₯^2 ) Now, 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Putting value of 𝑑𝑒/𝑑π‘₯ & 𝑑𝑣/𝑑π‘₯ π’…π’š/𝒅𝒙 = (𝒙 𝒄𝒐𝒔⁑𝒙 )^𝒙 (𝟏 βˆ’ 𝒙 𝒕𝒂𝒏⁑𝒙+π’π’π’ˆβ‘(𝒙 𝒄𝒐𝒔⁑𝒙 ) ) + (𝒙 π’”π’Šπ’β‘π’™ )^(𝟏/𝒙) ((〖𝒙 𝒄𝒐𝒕〗⁑〖𝒙 γ€— + 𝟏 βˆ’ π₯𝐨𝐠 (𝒙 π’”π’Šπ’β‘π’™ ))/𝒙^𝟐 )

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.