Given the equation:
\[x \cos(p + y) + \cos p \sin(p + y) = 0.\]
1. Rearrange to isolate the term with \( \sin(p+y) \):
\[\cos p \sin(p + y) = -x \cos(p + y).\]
2. Differentiate both sides with respect to \( x \), treating \( p \) as a constant:
\[\cos p \frac{d}{dx} \big[\sin(p + y)\big] = -\frac{d}{dx} \big[x \cos(p + y)\big].\]
3. Apply the chain rule on the left side and the product rule on the right side:
\[\cos p \cos(p + y) \frac{dy}{dx} = -\big[1 \cdot \cos(p + y) + x(-\sin(p + y))\frac{dy}{dx}\big].\]
4. Simplify the equation:
\[\cos p \cos(p + y) \frac{dy}{dx} = -\cos(p + y) + x \sin(p + y) \frac{dy}{dx}.\]
5. Group the \( \frac{dy}{dx} \) terms on one side:
\[\frac{dy}{dx} \big[\cos p \cos(p + y) - x \sin(p + y)\big] = -\cos(p + y).\]
6. Solve for \( \frac{dy}{dx} \):
\[\frac{dy}{dx} = \frac{-\cos(p + y)}{\cos p \cos(p + y) - x \sin(p + y)}.\]
Alternatively, if \( \cos(p+y) eq 0 \), we can simplify the original rearranged equation in step 1 and differentiate. From step 1, \( \cos p \sin(p + y) = -x \cos(p + y) \).
If \( \cos(p+y) eq 0 \), divide by \( \cos(p+y) \):
\[\cos p \tan(p+y) = -x.\]
Differentiate both sides with respect to \( x \):
\[\cos p \sec^2(p+y) \frac{dy}{dx} = -1.\]
Solve for \( \frac{dy}{dx} \):
\[\frac{dy}{dx} = \frac{-1}{\cos p \sec^2(p+y)} = \frac{-\cos p}{\sec^2(p+y)}.\]
Using \( \sec^2(p+y) = \frac{1}{\cos^2(p+y)} \):
\[\frac{dy}{dx} = -\cos p \cos^2(p+y).\]
Let's verify this with the result from step 6. From step 6:
\[\frac{dy}{dx} = \frac{-\cos(p + y)}{\cos p \cos(p + y) - x \sin(p + y)}.\]
From the original equation, \( x \cos(p + y) = -\cos p \sin(p + y) \).
Substitute \( x = \frac{-\cos p \sin(p + y)}{\cos(p+y)} \) into the denominator:
\[\cos p \cos(p + y) - \left(\frac{-\cos p \sin(p + y)}{\cos(p+y)}\right) \sin(p + y)\]
\[= \cos p \cos(p + y) + \frac{\cos p \sin^2(p + y)}{\cos(p+y)}\]
\[= \frac{\cos p \cos^2(p + y) + \cos p \sin^2(p + y)}{\cos(p+y)}\]
\[= \frac{\cos p (\cos^2(p + y) + \sin^2(p + y))}{\cos(p+y)}\]
\[= \frac{\cos p}{\cos(p+y)}.\]
So,
\[\frac{dy}{dx} = \frac{-\cos(p + y)}{\frac{\cos p}{\cos(p+y)}} = \frac{-\cos^2(p + y)}{\cos p}.\]
This matches the simplified result.
Proved.