Question 23 (B) - Differentiate (cos x)^x with respect to x [Video]

Question 23 (B) Differentiate the following function with respect to x:(cos 𝑥)^𝑥; (where ├ 𝑥∈(0,𝜋/2)).Let 𝑦 = 〖(𝑐𝑜𝑠⁡𝑥 ) 〗^𝑥 We use log differentiation Taking log both sides . log⁡𝑦 = log〖 (𝑐𝑜𝑠⁡𝑥 ) 〗^𝑥 𝒍𝒐𝒈⁡𝒚 = 𝒙 . 𝒍𝒐𝒈 (𝒄𝒐𝒔⁡𝒙 ) Differentiating both sides 𝑤.𝑟.𝑡.𝑥. (As 𝑙𝑜𝑔⁡(𝑎^𝑏 )=𝑏 . 𝑙𝑜𝑔⁡𝑎) 𝑑(log⁡𝑦 )/𝑑𝑥 = (𝑑(𝑥 . log⁡(cos⁡𝑥 ) ) )/𝑑𝑥 𝑑(log⁡𝑦 )/𝑑𝑦 . 𝑑𝑦/𝑑𝑥 = (𝑑(𝑥 . log⁡(cos⁡𝑥 ) ) )/𝑑𝑥 1/𝑦 (𝑑𝑦/𝑑𝑥) = (𝑑(𝑥 . log⁡(cos⁡𝑥 ) ) )/𝑑𝑥 1/𝑦 (𝑑𝑦/𝑑𝑥) = 𝑑𝑥/𝑑𝑥 log⁡〖( 𝑐𝑜𝑠 𝑥)〗+𝑥 (𝑑(𝑙𝑜𝑔⁡(𝑐𝑜𝑠⁡𝑥 ) ) )/𝑑𝑥 1/𝑦 (𝑑𝑦/𝑑𝑥) = log⁡〖(cos⁡𝑥)〗+𝑥 × 1/cos⁡𝑥 × (𝑐𝑜𝑠 𝑥)^′ 1/𝑦 (𝑑𝑦/𝑑𝑥) = log⁡〖(cos⁡𝑥)〗+𝑥/cos⁡𝑥 × (−sin⁡𝑥 ) Using product Rule As (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 1/𝑦 (𝑑𝑦/𝑑𝑥) = log⁡〖(cos⁡𝑥)〗−𝑥 tan⁡𝑥 𝑑𝑦/𝑑𝑥 = 𝑦[log⁡(cos⁡𝑥 )−𝑥 tan⁡𝑥 ] Putting value of 𝑦 𝒅𝒚/𝒅𝒙 = (𝐜𝐨𝐬⁡𝒙 )^𝒙 [𝒍𝒐𝒈⁡(𝒄𝒐𝒔⁡𝒙 )−𝒙 𝒕𝒂𝒏⁡𝒙 ]

Question 23 (B) - CBSE Class 12 Sample Paper for 2025 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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