Check Full Chapter Explained - Continuity and Differentiability - https://you.tube/Chapter-5-Class-12-Continuity

Slide1.JPG

Slide2.JPG
Slide3.JPG Slide4.JPG Slide5.JPG Slide6.JPG Slide7.JPG

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

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 )

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.