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





  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards


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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.