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

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

Β 

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

Ques 18 (MCQ) - If y = log(cos e^x), then dy/dx is: [Video] - Teachoo - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

part 2 - Question 18 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

Share on WhatsApp

Transcript

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)

Davneet Singh's photo - Co-founder, Teachoo

Made by

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