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

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

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Transcript

Question 18 If y = ๐‘™๐‘œ๐‘”โก(๐‘๐‘œ๐‘ โกใ€–๐‘’^๐‘ฅ ใ€— ), then ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ is : (a) ๐‘๐‘œ๐‘ ๐‘’^(๐‘ฅโˆ’1) (b) ๐‘’^(โˆ’๐‘ฅ) "cos" ๐‘’^๐‘ฅ (c) ๐‘’^๐‘ฅ ๐‘ ๐‘–๐‘›๐‘’^๐‘ฅ (d) ใ€–โˆ’๐‘’ใ€—^๐‘ฅ " tan" ใ€– ๐‘’ใ€—^๐‘ฅ Given ๐‘ฆ = ใ€–log ใ€—โก(cosโกใ€–๐‘’^๐‘ฅ ใ€— ) Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘(๐‘ฆ)/๐‘‘๐‘ฅ = ๐‘‘(ใ€–log ใ€—โก(cosโกใ€–๐‘’^๐‘ฅ ใ€— ) )/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/cosโกใ€–๐‘’^๐‘ฅ ใ€— . ๐‘‘(cosโกใ€–๐‘’^๐‘ฅ ใ€— )/๐‘‘๐‘ฅ = 1/cosโกใ€–๐‘’^๐‘ฅ ใ€— . (โˆ’sinโกใ€–๐‘’^๐‘ฅ ใ€— ) . ๐‘‘(๐‘’^๐‘ฅ )/๐‘‘๐‘ฅ = 1/cosโกใ€–๐‘’^๐‘ฅ ใ€— . (โˆ’sinโกใ€–๐‘’^๐‘ฅ ใ€— ) . ๐‘’^๐‘ฅ = (โˆ’sinโกใ€–๐‘’^๐‘ฅ ใ€—)/cosโกใ€–๐‘’^๐‘ฅ ใ€— . ๐‘’^๐‘ฅ = โˆ’tanโกใ€–๐‘’^๐‘ฅ ใ€— . ๐‘’^๐‘ฅ = โˆ’๐’†^๐’™ . ๐’•๐’‚๐’โกใ€–๐’†^๐’™ ใ€— So, the correct answer is (D)

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