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Given the equation: $x \cos(p+y) + \cos p \sin(p+y) = 0$. We want to find $\frac{dy}{dx}$.
Differentiating both sides of the equation with respect to $x$, we get: $$ \frac{d}{dx} [x \cos(p+y) + \cos p \sin(p+y)] = \frac{d}{dx} [0] $$ $$ \frac{d}{dx} [x \cos(p+y)] + \frac{d}{dx} [\cos p \sin(p+y)] = 0 $$
Using the product rule for the first term and the chain rule for both terms: $$ \left[ (1) \cos(p+y) + x (-\sin(p+y)) \frac{dy}{dx} \right] + \cos p \left[ \cos(p+y) \frac{dy}{dx} \right] = 0 $$ $$ \cos(p+y) - x \sin(p+y) \frac{dy}{dx} + \cos p \cos(p+y) \frac{dy}{dx} = 0 $$
Now, we isolate $\frac{dy}{dx}$: $$ \frac{dy}{dx} [ -x \sin(p+y) + \cos p \cos(p+y) ] = - \cos(p+y) $$ $$ \frac{dy}{dx} = \frac{- \cos(p+y)}{-x \sin(p+y) + \cos p \cos(p+y)} $$
From the original equation, we have $x \cos(p+y) = - \cos p \sin(p+y)$, so $x = - \frac{\cos p \sin(p+y)}{\cos(p+y)}$. Substituting this into the expression for $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{- \cos(p+y)}{- \left( - \frac{\cos p \sin(p+y)}{\cos(p+y)} \right) \sin(p+y) + \cos p \cos(p+y)} $$ $$ \frac{dy}{dx} = \frac{- \cos(p+y)}{\frac{\cos p \sin^2(p+y)}{\cos(p+y)} + \cos p \cos(p+y)} $$ $$ \frac{dy}{dx} = \frac{- \cos(p+y)}{\frac{\cos p \sin^2(p+y) + \cos p \cos^2(p+y)}{\cos(p+y)}} $$ $$ \frac{dy}{dx} = \frac{- \cos^2(p+y)}{\cos p (\sin^2(p+y) + \cos^2(p+y))} $$ $$ \frac{dy}{dx} = \frac{- \cos^2(p+y)}{\cos p (1)} $$ $$ \frac{dy}{dx} = \frac{- \cos^2(p+y)}{\cos p} $$
Multiplying both sides by $\cos p$, we get: $$ \cos p \frac{dy}{dx} = - \cos^2(p+y) $$
Final Answer: $\cos p \frac{dy}{dx} = - \cos^2(p+y)$