Find the dy/dx

y= sin²x/cos²x

d y dx=0

Explanation:

From the given

y=sin2x+cos2x

The right side of the equation is

=1

y=1

d y d x=d d x(1)=0

or we can do it this way.

y = sin2x+cos2x

d y d x=2

⋅(sinx)2−1

⋅

d d x(sinx)+2

(cosx)2−1 d d x(cosx)

d y d x=2⋅

sinx⋅cosx+2⋅cosx⋅(−sinx)

dydx=2⋅sinx⋅cosx−2⋅sinx⋅cosx

dydx=0

God blessI hope the explanation is useful.

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this