Question

Find nth derivative of cos x .cos 2x.cos 3x

Answer 1

Answer:

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) ]

I think it will help you

Answer 2

Given :  cosx cos2x cos3x

To Find :  nth Derivative

Solution:

y ​= cos2x . cos3x.cosx

cosA. cosB = (1/2)( cos(A + B) + Cos(A - B)

y = cos2x .(1/2)) (cos 4x + Cos2x)

y = (1/2)cos2xcos 4x + (1/2)Cos²2x

y =  (1/2)cos4xcos 2x + (1/2)Cos²2x

1 + cos2A = 2Cos²A

y = (1/2)(1/2)( cos6x + cos2x)  + (1/2)(1/2)(1 + cos4x )

y = 1/4  + (1/4)( cos2x + cos4x  + cos6x)

dⁿcos(ax+ b) = aⁿ cos( ax+ b + nπ/2)

yₙ = nth derivative =  (1/4)  (2ⁿcos(2x +  nπ/2)  + 4ⁿcos(4x +  nπ/2) + 6ⁿcos(6x +  nπ/2))

(1/4)  (2ⁿcos(2x +  nπ/2)  + 4ⁿcos(4x +  nπ/2) + 6ⁿcos(6x +  nπ/2))

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