This question is similar to Chapter 5 Class 12 Continuity and Differentiability - Ex 5.5
Please check the question here
CBSE Class 12 Sample Paper for 2025 Boards
Question 2
Question 3
Question 4
Question 5 Important
Question 6
Question 7
Question 8 Important
Question 9 Important
Question 10 Important
Question 11
Question 12 Important
Question 13 Important
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Question 15 Important
Question 16 Important
Question 17
Question 18
Question 19 [Assertion Reasoning] Important
Question 20 [Assertion Reasoning] Important
Question 21 Important
Question 22
Question 23 (A)
Question 23 (B) You are here
Question 24 (A)
Question 24 (B) Important
Question 25 Important
Question 26 Important
Question 27 Important
Question 28 (A)
Question 28 (B)
Question 29 (A) Important
Question 29 (B)
Question 30 Important
Question 31 (A) Important
Question 31 (B)
Question 32 Important
Question 33 Important
Question 34 (A)
Question 34 (B)
Question 35 (A)
Question 35 (B) Important
Question 36 (i) [Case Based]
Question 36 (ii)
Question 36 (iii) (A) Important
Question 36 (iii) (B) Important
Question 37 (i) [Case Based]
Question 37 (ii) Important
Question 37 (iii) (A) Important
Question 37 (iii) (B)
Question 38 (i) [Case Based] Important
Question 38 (ii) Important
CBSE Class 12 Sample Paper for 2025 Boards
Last updated at Oct. 3, 2024 by Teachoo
This question is similar to Chapter 5 Class 12 Continuity and Differentiability - Ex 5.5
Please check the question here
Question 23 (B) Differentiate the following function with respect to x:(cos 𝑥)^𝑥; (where ├ 𝑥∈(0,𝜋/2)).Let 𝑦 = 〖(𝑐𝑜𝑠𝑥 ) 〗^𝑥 We use log differentiation Taking log both sides . log𝑦 = log〖 (𝑐𝑜𝑠𝑥 ) 〗^𝑥 𝒍𝒐𝒈𝒚 = 𝒙 . 𝒍𝒐𝒈 (𝒄𝒐𝒔𝒙 ) Differentiating both sides 𝑤.𝑟.𝑡.𝑥. (As 𝑙𝑜𝑔(𝑎^𝑏 )=𝑏 . 𝑙𝑜𝑔𝑎) 𝑑(log𝑦 )/𝑑𝑥 = (𝑑(𝑥 . log(cos𝑥 ) ) )/𝑑𝑥 𝑑(log𝑦 )/𝑑𝑦 . 𝑑𝑦/𝑑𝑥 = (𝑑(𝑥 . log(cos𝑥 ) ) )/𝑑𝑥 1/𝑦 (𝑑𝑦/𝑑𝑥) = (𝑑(𝑥 . log(cos𝑥 ) ) )/𝑑𝑥 1/𝑦 (𝑑𝑦/𝑑𝑥) = 𝑑𝑥/𝑑𝑥 log〖( 𝑐𝑜𝑠 𝑥)〗+𝑥 (𝑑(𝑙𝑜𝑔(𝑐𝑜𝑠𝑥 ) ) )/𝑑𝑥 1/𝑦 (𝑑𝑦/𝑑𝑥) = log〖(cos𝑥)〗+𝑥 × 1/cos𝑥 × (𝑐𝑜𝑠 𝑥)^′ 1/𝑦 (𝑑𝑦/𝑑𝑥) = log〖(cos𝑥)〗+𝑥/cos𝑥 × (−sin𝑥 ) Using product Rule As (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 1/𝑦 (𝑑𝑦/𝑑𝑥) = log〖(cos𝑥)〗−𝑥 tan𝑥 𝑑𝑦/𝑑𝑥 = 𝑦[log(cos𝑥 )−𝑥 tan𝑥 ] Putting value of 𝑦 𝒅𝒚/𝒅𝒙 = (𝐜𝐨𝐬𝒙 )^𝒙 [𝒍𝒐𝒈(𝒄𝒐𝒔𝒙 )−𝒙 𝒕𝒂𝒏𝒙 ]