If xy=(x+y)^2 , then (d^2y)/(dx^2) is equal to:

The correct answer is Option (4) → 0

Expanding the right side:

    \( \displaystyle xy=x^2+2xy+y^2 \)

    \( \displaystyle x^2+xy+y^2=0 \)

Differentiating with respect to x:

    \( \displaystyle 2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0 \)

    \( \displaystyle \frac{dy}{dx}=-\frac{2x+y}{x+2y} \)

This equals \( \frac{y}{x} \), because

    \( \displaystyle \frac{y}{x}+\frac{2x+y}{x+2y} \)

    \( \displaystyle =\frac{y(x+2y)+x(2x+y)}{x(x+2y)} \)

    \( \displaystyle =\frac{2(x^2+xy+y^2)}{x(x+2y)}=0 \)

So \( \frac{dy}{dx}=\frac{y}{x} \). Differentiating again by the quotient rule:

    \( \displaystyle \frac{d^2y}{dx^2}=\frac{x\frac{dy}{dx}-y}{x^2} \)

    \( \displaystyle =\frac{x\cdot\frac{y}{x}-y}{x^2} \)

    \( \displaystyle =0 \)

Tip: For any relation of the form \( x^my^n=(x+y)^{m+n} \), \( \frac{dy}{dx}=\frac{y}{x} \). This question is the case m = n = 1, and then \( \frac{d^2y}{dx^2}=0 \) follows at once.

Note: It seems that there is a mistake in the original NTA paper, which prints the relation \( xy=(x+y)^2 \). It simplifies to \( x^2+xy+y^2=0 \), that is, \( \left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2=0 \), which holds for real x and y only at the point (0, 0), where \( \frac{dy}{dx} \) is not defined; so, strictly, y is not a differentiable function of x here. The question can still be solved because the usual working, the case m = n = 1 of \( x^my^n=(x+y)^{m+n} \), gives 0, the answer given by NTA.