If y = log⁡(cos⁡ e x ), then dy/dx is :

(a) cose (x-1)   (b) e (-x) cos e x
(c) e x sine x    (d) -e x tan e x

 

This question is inspired from Ex 5.4, 5 - Chapter 5 Class 12 - Continuity and Differentiability

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Question 18 If y = π‘™π‘œπ‘”β‘(π‘π‘œπ‘ β‘γ€–π‘’^π‘₯ γ€— ), then 𝑑𝑦/𝑑π‘₯ is : (a) π‘π‘œπ‘ π‘’^(π‘₯βˆ’1) (b) 𝑒^(βˆ’π‘₯) "cos" 𝑒^π‘₯ (c) 𝑒^π‘₯ 𝑠𝑖𝑛𝑒^π‘₯ (d) γ€–βˆ’π‘’γ€—^π‘₯ " tan" γ€– 𝑒〗^π‘₯ Given 𝑦 = γ€–log 〗⁑(cos⁑〖𝑒^π‘₯ γ€— ) Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑(𝑦)/𝑑π‘₯ = 𝑑(γ€–log 〗⁑(cos⁑〖𝑒^π‘₯ γ€— ) )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 1/cos⁑〖𝑒^π‘₯ γ€— . 𝑑(cos⁑〖𝑒^π‘₯ γ€— )/𝑑π‘₯ = 1/cos⁑〖𝑒^π‘₯ γ€— . (βˆ’sin⁑〖𝑒^π‘₯ γ€— ) . 𝑑(𝑒^π‘₯ )/𝑑π‘₯ = 1/cos⁑〖𝑒^π‘₯ γ€— . (βˆ’sin⁑〖𝑒^π‘₯ γ€— ) . 𝑒^π‘₯ = (βˆ’sin⁑〖𝑒^π‘₯ γ€—)/cos⁑〖𝑒^π‘₯ γ€— . 𝑒^π‘₯ = βˆ’tan⁑〖𝑒^π‘₯ γ€— . 𝑒^π‘₯ = βˆ’π’†^𝒙 . 𝒕𝒂𝒏⁑〖𝒆^𝒙 γ€— So, the correct answer is (D)

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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