2cos2x=y
dy/dx(2cos2x)=-4sin2x
d/du cosu= u'×cos'u=-u'×sinu
d/du sinu= u'×sin'u= u'×cosu
d/du tanu = u'×(1+tan²u)
d/du cotu = u'×-(1+cot²u)
proof for cosu :
cos2x= 1-2sin²x
or 2cosx²-1
or cos²x-sin²x
we will one of this fro example let's take the first one
d/dx 1-2sin²x=0-2×2(sinx)×cosx=-2sin2x
and we have 2 in the begining so its -4sin2x
