Question

If $x^p{y}^q=(x+y{)}^{p+q}$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{y}{x}$
• B. $\frac{py}{qx}$
• C. $\frac{x}{y}$
• D. $\frac{qy}{px}$

Answer 1

The correct option is A $\frac{y}{x}$

Given, $x^p{y}^q=(x+y{)}^{p+q}$
Taking $\log$ on both sides, we get
$p\log x+q\log y=(p+q)\log (x+y)$
$\Rightarrow \frac{p}{x}+\frac{q}{y}\frac{dy}{dx}=\frac{(p+q)}{(x+y)}(1+\frac{dy}{dx})$
$\Rightarrow (\frac{p}{x}-\frac{p+q}{x+y})=(\frac{p+q}{x+y}-\frac{q}{y})\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}$