This question is similar to Chapter 5 Class 12 Continuity and Differentiability - Ex 5.5

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https://www.teachoo.com/3677/701/Ex-5.5--11---Differentiate-(x-cos-x)x---(x-sin-x)1-x/category/Ex-5.5/

 

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Transcript

Question 23 (B) Differentiate the following function with respect to x:(cos π‘₯)^π‘₯; (where β”œ π‘₯∈(0,πœ‹/2)).Let 𝑦 = γ€–(π‘π‘œπ‘ β‘π‘₯ ) γ€—^π‘₯ We use log differentiation Taking log both sides . log⁑𝑦 = logγ€– (π‘π‘œπ‘ β‘π‘₯ ) γ€—^π‘₯ π’π’π’ˆβ‘π’š = 𝒙 . π’π’π’ˆ (𝒄𝒐𝒔⁑𝒙 ) Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏 )=𝑏 . π‘™π‘œπ‘”β‘π‘Ž) 𝑑(log⁑𝑦 )/𝑑π‘₯ = (𝑑(π‘₯ . log⁑(cos⁑π‘₯ ) ) )/𝑑π‘₯ 𝑑(log⁑𝑦 )/𝑑𝑦 . 𝑑𝑦/𝑑π‘₯ = (𝑑(π‘₯ . log⁑(cos⁑π‘₯ ) ) )/𝑑π‘₯ 1/𝑦 (𝑑𝑦/𝑑π‘₯) = (𝑑(π‘₯ . log⁑(cos⁑π‘₯ ) ) )/𝑑π‘₯ 1/𝑦 (𝑑𝑦/𝑑π‘₯) = 𝑑π‘₯/𝑑π‘₯ log⁑〖( π‘π‘œπ‘  π‘₯)γ€—+π‘₯ (𝑑(π‘™π‘œπ‘”β‘(π‘π‘œπ‘ β‘π‘₯ ) ) )/𝑑π‘₯ 1/𝑦 (𝑑𝑦/𝑑π‘₯) = log⁑〖(cos⁑π‘₯)γ€—+π‘₯ Γ— 1/cos⁑π‘₯ Γ— (π‘π‘œπ‘  π‘₯)^β€² 1/𝑦 (𝑑𝑦/𝑑π‘₯) = log⁑〖(cos⁑π‘₯)γ€—+π‘₯/cos⁑π‘₯ Γ— (βˆ’sin⁑π‘₯ ) Using product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 1/𝑦 (𝑑𝑦/𝑑π‘₯) = log⁑〖(cos⁑π‘₯)γ€—βˆ’π‘₯ tan⁑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑦[log⁑(cos⁑π‘₯ )βˆ’π‘₯ tan⁑π‘₯ ] Putting value of 𝑦 π’…π’š/𝒅𝒙 = (πœπ¨π¬β‘π’™ )^𝒙 [π’π’π’ˆβ‘(𝒄𝒐𝒔⁑𝒙 )βˆ’π’™ 𝒕𝒂𝒏⁑𝒙 ]

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo