Question

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

• A. $-\frac{y}{x}$
• B. $\frac{y}{x}$
• C. $\frac{x}{y}$
• D. None of these

Answer 1

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

We have $x^m{y}^n=(x+y{)}^{m+n}$.
Taking log on both sides, we get
$m\log x+n\log y=(m+n)\log (x+y)$
Differentiating both sides w.r.t. x, we get
$m\frac{1}{x}+n\frac{1}{y}\frac{dy}{dx}=\frac{m+n}{x+y}\frac{d}{dx}(x+y)$
or $(\frac{m}{x}+\frac{n}{y})\frac{dy}{dx}=\frac{m+n}{x+y}(1+\frac{dy}{dx})$
or $\{\frac{n}{y}-\frac{m+n}{x+y}\}\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}$
or $\left\{\frac{nx+ny-my-ny}{y(x+y)}\right\}\frac{dy}{dx}=\left\{\frac{mx+nx-mx-my}{(x+y)x}\right\}$
or $\frac{nx-my}{y(x+y)}\frac{dy}{dx}=\frac{nx-my}{(x+y)x}$
or $\frac{dy}{dx}=\frac{y}{x}$