Find-the-nth-derivative-of-sin-2-2x- Tinku Tara June 3, 2023 Differentiation 0 Comments FacebookTweetPin Question Number 8996 by Basant007 last updated on 11/Nov/16 Findthenthderivativeofsin22x Commented by FilupSmith last updated on 13/Nov/16 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)I′mnotsureifthereisageneralformfordnydxn Answered by 123456 last updated on 14/Nov/16 y=sin22xdydx=2sin4xd2ydx2=8cos4x=8sin(4x+π2)d3ydx3=−32sin4x=−32sin(4x+π)d4ydx4=−128cos4x=−128sin(4x+3π2)d5ydx5=512sin4x−−−−−−−1→2=2⋅40;02→8=2⋅41=23;π23→32=2⋅42=25;π4→128=2⋅43=27;3π2n→2⋅4n−1=2⋅22n−2=22n−1;(n−1)π2dnydxn={sin22xn=022n−1sin(4x+(n−1)π2)n>0 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-140064Next Next post: Question-8999 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.