Find the derivative of the following w.r.t x: 1)sin2x 2)sinmx cosnx Share with your friends Share 3 Ajanta Trivedi answered this ddxsin2x=2sinx*ddxsinx=2sinx*cosx=sin2x ddxsinmxcosnx=ddxsinmxcosnx+sinmx.ddx(cosnx)=msinm-1xddxsinxcosnx+sinmx.ncosn-1x.ddxcosx=msinm-1x.cosx.cosnx+n.sinmx.cosn-1x.-sinx=msinm-1x.cosn+1x-nsinm+1x.cosn-1x hope this helps you -5 View Full Answer Abantika Ghosh answered this 1) d/dx (sin2x) = 2sinx . d/dx (sinx)= -2sinxcosx2) d/dx(sinmxcosnx)= cosnx. d/dx (sinmx) + sinmx . d/dx (cosnx)= mcosnx .sinm-1x (cosx) - nsinmxcosn-1x (sinx)=mcosn+1xsinm-1x -- nsinm+1xcosn-1x = cosn-1xsinm-1x(mcos2x -- nsin2x) 0