Question

If $x^m{y}^n=(x+y{)}^{m+n}$ then $\frac{dy}{dx}$ is equal to

• A. $\frac{x+y}{xy}$
• B. $xy$
• C. $0$
• D. $\frac{y}{x}$

Answer 1

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

$x^m{y}^n=(x+y{)}^{m+n}$
Taking log both sides, we get,
$\Rightarrow m\log x+n\log y=(m+n)\log (x+y)$
Differentiating both sides w.r.t. $x$, we get
$\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}(1+\frac{dy}{dx})$
$\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}+\frac{m+n}{x+y}\frac{dy}{dx}$
$\frac{m}{x}-\frac{m+n}{x+y}=(\frac{m+n}{x+y}-\frac{n}{y})\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{my-nx}{my-nx}\frac{y}{x}$
$\frac{dy}{dx}=\frac{y}{x}$