Respuesta :

Answer:

sin²2θ. (cos θ sin θ). cos 2θ

Step-by-step explanation:

finding g'(x)

g'(x)

  • (x^n)' = nx^(n -1)

= 4 (cosθsinθ)³ . { cosθ. (sinθ)' + sinθ. (cosθ)' }

  • (cosθ)' = - sinθ
  • (sinθ)' = cosθ

= 4 (cosθsinθ)³ { cosθ. cos θ + sinθ.(-sin θ)}

= 4 (cosθsinθ)³{ cos²θ - sin²θ}

  • cos²θ - sin²θ = cos 2θ
  • 2sinθ cosθ = sin 2θ

= (4 cosθ sinθ)². (cosθ sinθ). { cos²θ - sin²θ}

= sin²2θ. (cos θ sin θ). cos 2θ