The derivative of $y$ with respect to $x$

Question

Question

Consider the following for the two (02) items that follow:

Let $(x + y)^{p + q} = x p y q$ , where $p, q$ are positive integers.

This question was previously asked in

NDA-I (Mathematics) Official Paper (Held On: 13 Apr, 2025)

Answer 1

Calculation:

Given,

$(x+y)^{p+q} = x^p\,y^q$

Differentiate implicitly w.r.t. $x$:

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

Rearrange to collect $\tfrac{dy}{dx}$:

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

Use $(x+y)^{p+q-1}=\frac{x^p\,y^q}{x+y}$ to simplify:

$\tfrac{dy}{dx} = \tfrac{y}{x}$

∴ $\displaystyle \frac{dy}{dx} = \frac{y}{x}$, independent of $p$ and $q$.

Hence, the correct answer is Option 4.

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