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If $x^m \cdot y^n=(x+y)^{m+n}$, then prove that $\frac{d y}{d x}$ is $\frac{\mathrm{y}}{\mathrm{x}}$
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Consider $x^m \cdot y^n=(x+y)^{m+n}$ $\Rightarrow m \log x+n \log y=(m+n) \log (x+y)$ Differentiating both sides, we get
$$
\begin{aligned}
&\frac{m}{x}+\frac{n}{y} \frac{d y}{d x}=\frac{m+n}{x+y}\left(1+\frac{d y}{d x}\right) \\
&\Rightarrow\left(\frac{m}{x}-\frac{m+n}{x+y}\right)=\left(\frac{m+n}{x+y}-\frac{n}{y}\right) \frac{d y}{d x} \\
&\Rightarrow \frac{m y-n x}{x(x+y)}=\left(\frac{m y-n x}{y(x+y)}\right) \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{y}{x}
\end{aligned}
$$
$$
\begin{aligned}
&\frac{m}{x}+\frac{n}{y} \frac{d y}{d x}=\frac{m+n}{x+y}\left(1+\frac{d y}{d x}\right) \\
&\Rightarrow\left(\frac{m}{x}-\frac{m+n}{x+y}\right)=\left(\frac{m+n}{x+y}-\frac{n}{y}\right) \frac{d y}{d x} \\
&\Rightarrow \frac{m y-n x}{x(x+y)}=\left(\frac{m y-n x}{y(x+y)}\right) \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{y}{x}
\end{aligned}
$$
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