Question

y=sinx·cosx 和y=cos(2-4x)

Answer 1

y'=(sinx·cosx)'=(sinx)'cosx+sinx(cosx)'=cos²x-sin²x=cos2x

y'=[cos(2-4x)]'=[cos(4x-2)]'=-4sin(4x-2)

y'=(3e^x+5x^2-x+1)'=3e^x+10x-1

y'=[(1nx)/x]=[(lnx)'x-x'(lnx)]/ln²x=(1-lnx)/ln²x