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If $x y \neq 0, x+y \neq 0$ and $x^m y^n=(x+y)^{m+n}$, where $m, n \notin N$, then $\frac{d y}{d x}$ is equal to
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The correct answer is:
$\frac{y}{x}$
Given,
$x^m y^n=(x+y)^{m+n}$
On taking log on both sides, we get
$m \log x+n \log y=(m+n) \log (x+y)$
On differentiating w.r.t. $x$, we get
$\begin{array}{rlrl} & & \frac{m}{x}+\frac{n}{y} \frac{d y}{d x} & =\frac{m+n}{x+y}\left(1+\frac{d y}{d x}\right) \\ \Rightarrow & \frac{m}{x}-\frac{m+n}{x+y} & =\frac{d y}{d x}\left(\frac{m+n}{x+y}-\frac{n}{y}\right) \\ \Rightarrow & & \frac{m y-n x}{x(x+y)} & =\frac{d y}{d x}\left[\frac{m y-n x}{(x+y) y}\right] \\ \Rightarrow & & \frac{d y}{d x} & =\frac{y}{x}\end{array}$
$x^m y^n=(x+y)^{m+n}$
On taking log on both sides, we get
$m \log x+n \log y=(m+n) \log (x+y)$
On differentiating w.r.t. $x$, we get
$\begin{array}{rlrl} & & \frac{m}{x}+\frac{n}{y} \frac{d y}{d x} & =\frac{m+n}{x+y}\left(1+\frac{d y}{d x}\right) \\ \Rightarrow & \frac{m}{x}-\frac{m+n}{x+y} & =\frac{d y}{d x}\left(\frac{m+n}{x+y}-\frac{n}{y}\right) \\ \Rightarrow & & \frac{m y-n x}{x(x+y)} & =\frac{d y}{d x}\left[\frac{m y-n x}{(x+y) y}\right] \\ \Rightarrow & & \frac{d y}{d x} & =\frac{y}{x}\end{array}$
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