If $x^my^n=(x+y)^{m+n}$, then $\frac{d^2y}{dx^2}$ is equal to:

0

The correct answer is Option (4) → 0

Given: $x^{m}y^{n}=(x+y)^{m+n}$

Taking logarithm on both sides:

$m\log x + n\log y = (m+n)\log(x+y)$

Differentiating w.r.t. $x$:

$\frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = (m+n)\frac{1+\frac{dy}{dx}}{x+y}$

Rearranging terms and simplifying gives:

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

Differentiating again:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{y}{x}\right)=\frac{x\frac{dy}{dx}-y}{x^2}$

Substitute $\frac{dy}{dx}=\frac{y}{x}$:

$\frac{d^2y}{dx^2}=\frac{x\left(\frac{y}{x}\right)-y}{x^2}=0$

$\frac{d^2y}{dx^2}=0$