Question

If $x^p{y}^q=(x+y{)}^{p+q}$ then prove that $\frac{dy}{dx}=\frac{y}{x}$.

Answer 1

Given,

$x^p{y}^q=(x+y{)}^{p+q}$

Now taking logarithm both sides we get,

or, $p\log x+q\log y=(p+q)\log (x+y)$

Now differentiating both sides with respect to $x$ we get,

$\frac{p}{x}+\frac{q}{y}\frac{dy}{dx}=\frac{(p+q)}{x+y}(1+\frac{dy}{dx})$

or, $\frac{p}{x}-\frac{p+q}{x+y}=(\frac{p+q}{x+y}-\frac{q}{y})\frac{dy}{dx}$

or, $\frac{py-qx}{x(x+y)}=\frac{py-qx}{y(x+y)}\frac{dy}{dx}$

or, $\frac{dy}{dx}=\frac{y}{x}$.