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Question Number 8996 by Basant007 last updated on 11/Nov/16
Find the nth derivative of sin^2 2x
Findthenthderivativeofsin22x
Commented by FilupSmith last updated on 13/Nov/16
y=sin^2 (2x)  u=sin(2x) ⇒ du=2cos(2x)dx  dx=(1/(2cos(2x)))du  (dy/dx)=(d/dx)(u^2 )  (dy/dx)=(d/((1/(2cos(2x)))du))(u^2 )  (dy/dx)=2cos(2x)(d/du)(u^2 )  (dy/dx)=4cos(2x)u  (dy/dx)=4cos(2x)sin(2x)  cos(x)sin(x)=(1/2)sin(2x)  (dy/dx)=2sin(4x)     (d^2 y/dx^2 )=8cos(4x)     (d^3 y/dx^3 )=−32sin(4x)     (d^4 y/dx^4 )=−128cos(4x)     I′m not sure if there is a general form  for  (d^n y/dx^n )
y=sin2(2x)u=sin(2x)du=2cos(2x)dxdx=12cos(2x)dudydx=ddx(u2)dydx=d12cos(2x)du(u2)dydx=2cos(2x)ddu(u2)dydx=4cos(2x)udydx=4cos(2x)sin(2x)cos(x)sin(x)=12sin(2x)dydx=2sin(4x)d2ydx2=8cos(4x)d3ydx3=32sin(4x)d4ydx4=128cos(4x)Imnotsureifthereisageneralformfordnydxn
Answered by 123456 last updated on 14/Nov/16
y=sin^2 2x  (dy/dx)=2sin 4x  (d^2 y/dx^2 )=8cos 4x=8sin(4x+(π/2))  (d^3 y/dx^3 )=−32sin 4x=−32sin(4x+π)  (d^4 y/dx^4 )=−128cos 4x=−128sin(4x+((3π)/2))  (d^5 y/dx^5 )=512sin 4x  −−−−−−−  1→2=2∙4^0 ;0  2→8=2∙4^1 =2^3 ;(π/2)   3→32=2∙4^2 =2^5 ;π  4→128=2∙4^3 =2^7 ;((3π)/2)  n→2∙4^(n−1) =2∙2^(2n−2) =2^(2n−1) ;(((n−1)π)/2)  (d^n y/dx^n )= { ((sin^2 2x),(n=0)),((2^(2n−1) sin (4x+(((n−1)π)/2))),(n>0)) :}
y=sin22xdydx=2sin4xd2ydx2=8cos4x=8sin(4x+π2)d3ydx3=32sin4x=32sin(4x+π)d4ydx4=128cos4x=128sin(4x+3π2)d5ydx5=512sin4x12=240;028=241=23;π2332=242=25;π4128=243=27;3π2n24n1=222n2=22n1;(n1)π2dnydxn={sin22xn=022n1sin(4x+(n1)π2)n>0

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