PLS HELP I NEED TO TURN THIS IN SOON

I am pretty sure 40%

Step-by-step explanation:

if not srry

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