If $p+q=10$, then what is \(\frac{dy}{dx}\) equal to?

Calculation:

Given,

\( (x + y)^{p+q} = x^p\,y^q \) and \( p + q = 10 \).

Differentiate both sides with respect to \(x\) implicitly:

\(\frac{d}{dx}\bigl((x+y)^{p+q}\bigr) = \frac{d}{dx}\bigl(x^p y^q\bigr) \)

Left side:

\((p+q)\,(x+y)^{p+q-1}\bigl(1 + \tfrac{dy}{dx}\bigr) \)

Right side (product rule):

\(p\,x^{p-1}y^q \;+\; q\,x^p y^{q-1}\,\tfrac{dy}{dx} \)

Rearrange to collect \( \tfrac{dy}{dx} \) terms and use

\( (x+y)^{p+q} = x^p y^q \implies (x+y)^{p+q-1} = \tfrac{x^p y^q}{x+y} \).

After cancellation of the common factor \( p\,y - q\,x \), you obtain:

∴ \( \frac{dy}{dx} = \frac{y}{x} \)

Hence, the correct answer is Option 1.