Question

find the nth differential coefficient of cos power 4x​

Answer 1

Answer:

amm...........Method-1:

y=sin4x=(eix+e−ix2i)4=116(eix+e−ix)4  

⟹y=116(e4ix+e−4ix−4e2ix−4e−2ix+6)  

dnydxn=116((4i)ne4ix+(−4i)ne−4ix−4(2i)ne2ix−4(−2i)ne−2ix)  

=22n−4ine4ix+(−i)n22n−4e−4ix−2n−2ine2ix−(−i)n2n−2e−2ix  

Method-2:

y=sin4x=(1−cos2x2)2=14⋅[1+cos22x−2cos2x]  

=14⋅[1+1+cos4x2−2cos2x]  

⟹y=18⋅[3+cos4x−4cos2x]  

∴y′n=18⋅[4n⋅cos(4x+nπ2)−4⋅2n⋅cos(2x+nπ2)]

hope helpful!!