Find \( \frac{\partial w}{\partial x} \), where \( w=\cos z-\cos 3 x \cos 6 y+\sin 3 x \sin 6 y \) and \( z=3 x+6 y \) \( \frac{\partial w}{\partial x}=\square \)

To find \( \frac{\partial w}{\partial x} \) given \( w=\cos z-\cos 3 x \cos 6 y+\sin 3 x \sin 6 y \) and \( z=3 x+6 y \), we will use the chain rule and the product rule for differentiation. First, differentiate \( w \) with respect to \( x \): 1. Differentiate \( \cos z \): \[ \frac{\partial}{\partial x} \cos z = -\sin z \cdot \frac{\partial z}{\partial x} = -\sin(3x + 6y) \cdot 3. \] 2. Differentiate \( -\cos 3x \cos 6y \): \[ \frac{\partial}{\partial x}(-\cos 3x \cos 6y) = 3\sin 3x \cos 6y. \] 3. Differentiate \( \sin 3x \sin 6y \): \[ \frac{\partial}{\partial x}(\sin 3x \sin 6y) = 3\cos 3x \sin 6y. \] Now, combine these results: \[ \frac{\partial w}{\partial x} = -3\sin(3x + 6y) + 3\sin 3x \cos 6y + 3\cos 3x \sin 6y. \] Thus, we can factor out the 3: \[ \frac{\partial w}{\partial x} = 3\left( \sin 3x \cos 6y + \cos 3x \sin 6y - \sin(3x + 6y) \right). \] So the answer is: \[ \frac{\partial w}{\partial x} = 3\left( \sin 3x \cos 6y + \cos 3x \sin 6y - \sin(3x + 6y) \right). \]