If sin(x + y) + cos(2x + 2y) = ln(3x + 3y),th | vedclass

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(B) Given the equation: $\sin(x + y) + \cos(2x + 2y) = \ln(3(x + y))$.Differentiating both sides with respect to $x$ using the chain rule:$\cos(x + y)

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$-1$

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(B) Given the equation: $\sin(x + y) + \cos(2x + 2y) = \ln(3(x + y))$.Differentiating both sides with respect to $x$ using the chain rule:$\cos(x + y) \cdot (1 + \frac{dy}{dx}) - \sin(2x + 2y) \cdot (2 + 2\frac{dy}{dx}) = \frac{1}{3(x + y)} \cdot 3(1 + \frac{dy}{dx})$.Simplifying the right side:$\cos(x + y) \cdot (1 + \frac{dy}{dx}) - 2\sin(2x + 2y) \cdot (1 + \frac{dy}{dx}) = \frac{1}{x + y} \cdot (1 + \frac{dy}{dx})$.Rearranging the terms:$(1 + \frac{dy}{dx}) [\cos(x + y) - 2\sin(2x + 2y) - \frac{1}{x + y}] = 0$.Since the term in the square bracket is not identically zero,we must have:$1 + \frac{dy}{dx} = 0$.Therefore,$\frac{dy}{dx} = -1$.

If $\sin(x + y) + \cos(2x + 2y) = \ln(3x + 3y)$,then $\frac{dy}{dx}$ is

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