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If $x^{p}+y^{q}=(x+y)^{p+q}$, then $\frac{d y}{d x}$ is
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Verified Answer
The correct answer is:
$\frac{y}{x}$
If $x^{p}+y^{q}=(x+y)^{p+q}$
Taking log on both sides, $p \log x+q \log y=(p+q) \log (x+y)$
On differentiating w.r.t $x$, we get $\frac{p}{x}+\frac{q}{y} \cdot \frac{d y}{d x}=\frac{(p+q)}{(x+y)}\left(1+\frac{d y}{d x}\right)$
$$
\left\{\frac{p}{x}-\frac{p+q}{x+y}\right\}=\left\{\frac{p+q}{x+y}-\frac{q}{y}\right\} \frac{d y}{d x}
$$
$\left\{\frac{p x+p y-p x-q x}{x(x+y)}\right\}=\left\{\frac{p y+q y-q x-q y}{y(x+y)}\right\} \frac{d y}{d x}$
$\Rightarrow \frac{(p y-q x)}{x}=\frac{(p y-q x)}{y} \cdot \frac{d y}{d x}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{y}{x}$
Taking log on both sides, $p \log x+q \log y=(p+q) \log (x+y)$
On differentiating w.r.t $x$, we get $\frac{p}{x}+\frac{q}{y} \cdot \frac{d y}{d x}=\frac{(p+q)}{(x+y)}\left(1+\frac{d y}{d x}\right)$
$$
\left\{\frac{p}{x}-\frac{p+q}{x+y}\right\}=\left\{\frac{p+q}{x+y}-\frac{q}{y}\right\} \frac{d y}{d x}
$$
$\left\{\frac{p x+p y-p x-q x}{x(x+y)}\right\}=\left\{\frac{p y+q y-q x-q y}{y(x+y)}\right\} \frac{d y}{d x}$
$\Rightarrow \frac{(p y-q x)}{x}=\frac{(p y-q x)}{y} \cdot \frac{d y}{d x}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{y}{x}$
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