If $x^m \cdot y^n=(x+y)^{m+n}$, then $\frac{d y}{d x}$ is

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Aicharya · Accepted Answer

$x^m \cdot y^n=(x+y)^{m+n} \Rightarrow m \ln x+n \ln y=(m+n) \ln (x+y)$ $	herefore \frac{m}{x}+\frac{n}{y} \frac{d y}{d x}=\frac{m+n}{x+y}\left(1+\frac{d y}{d x}ight) \Rightarrow\left(\frac{m}{x}-\frac{m+n}{x+y}ight)=\left(\frac{m+n}{x+y}-\frac{n}{y}ight) \frac{d y}{d x}$ $\quad \frac{m y-n x}{x(x+y)}=\left(\frac{m y-n x}{y(x+y)}ight) \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{y}{x}$

If $x^m \cdot y^n=(x+y)^{m+n}$, then $\frac{d y}{d x}$ is

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