Question
If $x^p y^q=(x+y)^{p+q}$ then prove that $\frac{d y}{d x}=\frac{y}{x}$

Answer

$x^py^{q }= (x + y)^{p+q}$
Taking $\log$ both side
$p \log x + q \log y = (p + q) \log (x + y)$
Differentiating $w.r.t. x$
$ \frac{p}{x}+\frac{q}{y} \frac{d y}{d x}=\frac{p+q}{x+y}+\left(\frac{p+q}{x+y}\right) \frac{d y}{d x}$
$ \frac{q}{y} \frac{d y}{d x}-\left(\frac{p+q}{x+y}\right) \frac{d y}{d x}=\frac{p+q}{x+y}-\frac{p}{x}$
$ \left(\frac{q}{y}-\frac{p+q}{x+y}\right) \frac{d y}{d x}=\left(\frac{p+q}{x+y}-\frac{p}{x}\right)$
$ \left(\frac{q x-p y}{y}\right) \frac{d y}{d x}=\left(\frac{q x-p y}{x}\right)$
$ \frac{1}{y} \frac{d y}{d x}=\frac{1}{x}$
$ \frac{d y}{d x}=\frac{y}{x}$

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