JEE Main · 2006 · 30 Apr 2006 · Offline · Differentiation

Take logarithms of $x^m{y}^n=(x+y{)}^{m+n}$ to get $m\ln x+n\ln y=(m+n)\ln (x+y)$. Differentiating with respect to $x$ gives $\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right)$. Gathering the $\frac{dy}{dx}$ terms on one side, $\left(\frac{n}{y}-\frac{m+n}{x+y}\right)\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}$. Both sides reduce to the same factor: the left becomes $\frac{nx-my}{y(x+y)}\frac{dy}{dx}$ and the right becomes $\frac{nx-my}{x(x+y)}$. Cancelling the common $nx-my$ and $x+y$ leaves $\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}$, so $\frac{dy}{dx}=\frac{y}{x}$.