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Find the solution for Nth derivative of cosx*cos3x*cos5x

Praneet , 6 Years ago
Grade 12th pass
anser 1 Answers
Arun

Last Activity: 6 Years ago

cosx cos2x cos3x 
= 1/2 cos2x [ 2 cosx cos3x] [ cosC + cosD = 2 cos(C+D)/2 cos(C- D)/2 ] 
=1/2 cos2x [ cos4x + cos2x] 
= 1/4 [ 2 cos2x cos4x + 2 cos^22x ] 
= 1/4 [ cos6x + cos 2x + ( 1+ cos4x)] [ as cos2x = 2 cos^2x - 1 ] 
=1/4 cos6x +1/4 cos2x + 1/4 + 1/4 cos4x  
now D^n cos(ax+ b) = a^n cos( ax+ b+ npi/2) 
so D^n [cosx cos2x cos3x = 1/4D^n[ cos2x+ cos4x+ cos6x ] [ d/dx 1= 0 ] 
= 1/4[ 2^ncos(2x+npi/2)+4^n cos(4x+npi/2) + 6^n cos(6x+ npi/2) ]
 

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