Grade 12th passDifferential Calculus
How to get the nth order derivative of y=cos3x*cos2x
How to get the nth order derivative of y=cos3x*cos2x
2 Answers
Saurabh Koranglekar
Dear student
Please find the link below
https://www.askiitians.com/forums/Trigonometry/find-the-nth-derivative-of-cos2x-cos3x-please_220963.htm
Regards
Please find the link below
https://www.askiitians.com/forums/Trigonometry/find-the-nth-derivative-of-cos2x-cos3x-please_220963.htm
Regards
Arun
Dear Rishika
cos x cos 2x cos 3x = cos 2x [ cos 4x + cos 2x ] /2
=(1/2)[ cos 4x cos 2x + cos^2 2x]
=(1/4) [cos 6x + cos 2x + ( 1 + cos 4x)]
= (1/4) [ 1 + cos 2x + cos 4x + cos 6x ]
Now we have D^n ( cos (ax + b) = a^n cos ( n pi/2 + ax + b).
Thus, D^n( cos x cos 2x cos 3x ) = D^n [ (1/4) { 1 + cos 2x + cos 4x + cos 6x}]
= (1/4) [ 2^n cos(n pi/2 + 2x) + 4^n cos ( n pi/2 + 4x) + 6^n cos( n pi/2 + 6x) ]
=(1/2)[ cos 4x cos 2x + cos^2 2x]
=(1/4) [cos 6x + cos 2x + ( 1 + cos 4x)]
= (1/4) [ 1 + cos 2x + cos 4x + cos 6x ]
Now we have D^n ( cos (ax + b) = a^n cos ( n pi/2 + ax + b).
Thus, D^n( cos x cos 2x cos 3x ) = D^n [ (1/4) { 1 + cos 2x + cos 4x + cos 6x}]
= (1/4) [ 2^n cos(n pi/2 + 2x) + 4^n cos ( n pi/2 + 4x) + 6^n cos( n pi/2 + 6x) ]
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