Ex 5.5, 11 - Chapter 5 Class 12 Continuity and Differentiability (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 5 Class 12 Continuity and Differentiability
Ex 5.1, 13
Ex 5.1, 16
Ex 5.1, 18 Important
Ex 5.1, 28 Important
Ex 5.1, 30 Important
Ex 5.1, 34 Important
Ex 5.2, 5
Ex 5.2, 9 Important
Ex 5.2, 10 Important
Ex 5.3, 10 Important
Ex 5.3, 14
Example 29 Important
Example 30 Important
Ex 5.5,6 Important
Ex 5.5, 7 Important
Ex 5.5, 11 Important You are here
Ex 5.5, 16 Important
Ex 5.6, 7 Important
Ex 5.6, 11 Important
Example 38 Important
Ex 5.7, 14 Important
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Example 39 (i)
Example 40 (i)
Example 42 Important
Misc 6 Important
Misc 15 Important
Misc 16 Important
Misc 22 Important
Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 16, 2024 by Teachoo
Ex 5.5, 11 Differentiate the functions in, 〖(𝑥 𝑐𝑜𝑠𝑥 ) 〗^𝑥 + 〖(𝑥 𝑠𝑖𝑛𝑥 ) 〗^(1/𝑥)𝑦 = 〖(𝑥 𝑐𝑜𝑠𝑥 ) 〗^𝑥 + 〖(𝑥 𝑠𝑖𝑛𝑥 ) 〗^(1/𝑥) Let 𝑢 = 〖(𝑥 𝑐𝑜𝑠𝑥 ) 〗^𝑥 , 𝑣 = 〖(𝑥 𝑠𝑖𝑛𝑥 ) 〗^(1/𝑥) 𝑦 = 𝑢+𝑣 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦/𝑑𝑥 = (𝑑 (𝑢 + 𝑣))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑑𝑢/𝑑𝑥 + 𝑑𝑣/𝑑𝑥 Calculating 𝒅𝒖/𝒅𝒙 𝑢 =〖 (𝑥 𝑐𝑜𝑠𝑥 ) 〗^𝑥 Taking log both sides . log𝑢 = log〖 (𝑥 𝑐𝑜𝑠𝑥 ) 〗^𝑥 log𝑢 = 𝑥 . log (𝑥 𝑐𝑜𝑠𝑥 ) Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑(log𝑢 )/𝑑𝑥 = (𝑑(𝑥 . log(𝑥 cos𝑥 ) ) )/𝑑𝑥 𝑑(log𝑢 )/𝑑𝑢 . 𝑑𝑢/𝑑𝑥 = (𝑑(𝑥 . log(𝑥 cos𝑥 ) ) )/𝑑𝑥 1/𝑢 (𝑑𝑢/𝑑𝑥) = (𝑑(𝑥 . log(𝑥 cos𝑥 ) ) )/𝑑𝑥 (As 𝑙𝑜𝑔(𝑎^𝑏 )=𝑏 . 𝑙𝑜𝑔𝑎) Using product Rule As (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 1/𝑢 (𝑑𝑢/𝑑𝑥) = 𝑑𝑥/𝑑𝑥 log〖(𝑥 𝑐𝑜𝑠 𝑥)〗+𝑥 (𝑑(𝑙𝑜𝑔(𝑥 𝑐𝑜𝑠𝑥 ) ) )/𝑑𝑥 1/𝑢 (𝑑𝑢/𝑑𝑥) = log〖(𝑥 cos𝑥)〗+𝑥/(𝑥 cos𝑥 ) × (𝑥 𝑐𝑜𝑠 𝑥)^′ 1/𝑢 (𝑑𝑢/𝑑𝑥) = log〖(𝑥 cos𝑥)〗+1/cos𝑥 × (1.cos𝑥+𝑥(−sin𝑥 )) 1/𝑢 (𝑑𝑢/𝑑𝑥) = log〖(𝑥 cos𝑥)〗+((cos𝑥 − 𝑥 sin𝑥 ))/cos𝑥 1/𝑢 (𝑑𝑢/𝑑𝑥) = log〖(𝑥 cos𝑥)〗+cos𝑥/cos𝑥 −(𝑥 sin𝑥)/cos𝑥 1/𝑢 (𝑑𝑢/𝑑𝑥) = log〖(𝑥 cos𝑥)〗+1−𝑥 𝑡𝑎𝑛 𝑥 𝑑𝑢/𝑑𝑥 = u (1−𝑥 tan𝑥+log〖(𝑥 cos𝑥 〗)) Putting value of 𝑢 𝑑𝑢/𝑑𝑥 = (𝑥 cos𝑥 )^𝑥 (1−𝑥 tan𝑥+𝒍𝒐𝒈(𝒙 𝒄𝒐𝒔𝒙 ) ) Calculating 𝒅𝒗/𝒅𝒙 𝑣=〖(𝑥 𝑠𝑖𝑛𝑥 ) 〗^(1/𝑥) Taking log both sides log𝑣=log〖 〖(𝑥 𝑠𝑖𝑛𝑥 ) 〗^(1/𝑥) 〗 log𝑣= 1/𝑥 log (𝑥 𝑠𝑖𝑛𝑥 ) Differentiating both sides 𝑤.𝑟.𝑡.𝑥. (𝑑(log𝑣))/𝑑𝑥 = 𝑑(1/𝑥 " . " )/𝑑𝑥 (𝑑(log𝑣))/𝑑𝑣 . 𝑑𝑣/𝑑𝑥 = (1/𝑥 .log(𝑥 sin𝑥 ) )^′ 1/𝑣 × 𝑑𝑣/𝑑𝑥 = (1/𝑥 .log(𝑥 sin𝑥 ) )^′ 1/𝑣 × 𝑑𝑣/𝑑𝑥 = (((log〖(𝑥 sin𝑥)〗 )^′ 𝑥 + 𝑥^′ log〖(𝑥 sin𝑥)〗)/𝑥^2 ) Using quotient rule (𝑢/𝑣)^′ = (𝑢^′ 𝑣 − 𝑣^′ 𝑢)/𝑣^2 Where u = log (x sin x) , v = x 1/𝑣 × 𝑑𝑣/𝑑𝑥 = 1/x^2 ((log〖(𝑥 sin𝑥)〗 )^′ 𝑥 + log〖(𝑥 sin𝑥)〗 ) 1/𝑣 × 𝑑𝑣/𝑑𝑥 = 1/x^2 ([1/(𝑥 sin𝑥 ) \ ×(𝑥 sin𝑥 )^′ ]𝑥 + log〖(𝑥 sin𝑥)〗 ) 1/𝑣 × 𝑑𝑣/𝑑𝑥 = 1/x^2 (1/sin𝑥 \ ×(𝑥 sin𝑥 )^′+ log〖(𝑥 sin𝑥)〗 ) 1/𝑣 × 𝑑𝑣/𝑑𝑥 = 1/𝑥^2 (((1 . 𝑠𝑖𝑛𝑥 + 𝑥 𝑐𝑜𝑠𝑥))/𝑠𝑖𝑛𝑥 \ + 𝑙𝑜𝑔〖(𝑥 𝑠𝑖𝑛𝑥)〗 ) 1/𝑣 × 𝑑𝑣/𝑑𝑥 = 1/𝑥^2 (((𝑠𝑖𝑛𝑥 + 𝑥 𝑐𝑜𝑠𝑥))/𝑠𝑖𝑛𝑥 \ + 𝑙𝑜𝑔〖(𝑥 𝑠𝑖𝑛𝑥)〗 ) 1/𝑣 × 𝑑𝑣/𝑑𝑥 = 1/𝑥^2 (1+𝑥 cot𝑥+ 𝑙𝑜𝑔〖(𝑥 𝑠𝑖𝑛𝑥)〗 ) 𝑑𝑣/𝑑𝑥 = 𝑣((1 + 〖𝑥 𝑐𝑜𝑡〗〖𝑥 〗 − 𝑙𝑜𝑔〖 (𝑥 𝑠𝑖𝑛𝑥 ) 〗)/𝑥^2 ) 𝑑𝑣/𝑑𝑥 = (𝑥 𝑠𝑖𝑛𝑥 )^(1/𝑥) ((1 + 〖𝑥 𝑐𝑜𝑡〗〖𝑥 〗− 𝑙𝑜𝑔 (𝑥 𝑠𝑖𝑛𝑥 ))/𝑥^2 ) Now, 𝑑𝑦/𝑑𝑥 = 𝑑𝑢/𝑑𝑥 + 𝑑𝑣/𝑑𝑥 Putting value of 𝑑𝑢/𝑑𝑥 & 𝑑𝑣/𝑑𝑥 𝒅𝒚/𝒅𝒙 = (𝒙 𝒄𝒐𝒔𝒙 )^𝒙 (𝟏 − 𝒙 𝒕𝒂𝒏𝒙+𝒍𝒐𝒈(𝒙 𝒄𝒐𝒔𝒙 ) ) + (𝒙 𝒔𝒊𝒏𝒙 )^(𝟏/𝒙) ((〖𝒙 𝒄𝒐𝒕〗〖𝒙 〗 + 𝟏 − 𝐥𝐨𝐠 (𝒙 𝒔𝒊𝒏𝒙 ))/𝒙^𝟐 )