Question

Plz help me to answer

Answer 1

Answer:

for 3, 4 or both?

4. is probably zero

Step-by-step explanation:

Wait is 4 a question?

Answer 2

$cos^2 x + \cos^2 y = \cos(2x + 2y)\\\\\implies \dfrac{d}{dx} (\cos^2 x + \cos^2 y) = \dfrac{d}{dx} \cos(2x+2y)\\\\\implies -2\cos x \sin x -2 \cos y \sin y \dfrac{dy}{dx}=-\sin(2x+2y) \left(2+2\dfrac{dy}{dx} \right) \\\\\implies -2\cos x \sin x -2 \cos y \sin y \dfrac{dy}{dx}=-2\sin(2x+2y) -2\sin(2x+2y)\dfrac{dy}{dx} \right) \\\\\implies 2 \sin(2x+2y) \dfrac{dy}{dx} - 2 \cos y \sin y \dfrac{dy}{dx} = 2 \cos x \sin x-2\sin(2x+2y)$

$\implies \sin(2x+2y) \dfrac{dy}{dx} - \cos y \sin y \dfrac{dy}{dx} = \cos x \sin x-\sin(2x+2y)\\\\\implies \dfrac{dy}{dx} [\sin(2x+2y) - \cos y \sin y] = \cos x \sin x-\sin(2x+2y)\\\\\implies \dfrac{dy}{dx} [\sin(2x+2y) - \cos y \sin y] = \cos x \sin x-\sin(2x+2y)\\\\\implies \dfrac{dy}{dx} = \dfrac{\cos x \sin x-\sin(2x+2y)}{\sin(2x+2y) - \cos y \sin y}\\\\\implies y'(x) = \dfrac{\cos x \sin x-\sin(2x+2y)}{\sin(2x+2y) - \cos y \sin y}$