Question

If $x^m{y}^n=(x+y{)}^{m+n}$ then $y^{\prime }=$

• A. $\frac{-y}{x}$
• B. $\frac{x}{y}$
• C. $\frac{-x}{y}$
• D. $\frac{y}{x}$

Answer 1

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

Given $x^m{y}^n=(x+y{)}^{m+n}$

Taking log on both sides we get

$log{x}^m{y}^n=log(x+y{)}^{m+n}$

$\Rightarrow log{x}^m+log{y}^n=(m+n)log(x+y)$

$\Rightarrow mlogx+nlogy=(m+n)log(x+y)$

Differentiating above with respect to $x$, we get

$\Rightarrow \frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{(m+n)}{(x+y)}\times (1+\frac{dy}{dx})$

$\Rightarrow \frac{n}{y}\frac{dy}{dx}-\left(\frac{m+n}{x+y}\right)\times \frac{dy}{dx}=\left(\frac{m+n}{x+y}\right)-\frac{m}{x}$

$\Rightarrow (\frac{n}{y}-\frac{m+n}{x+y})\frac{dy}{dx}=\frac{x(m+n)-m(x+y)}{x(x+y)}$

$\Rightarrow \left(\frac{nx+ny-my-ny}{y(x+y)}\right)\frac{dy}{dx}=\frac{xm+xn-mx-my}{x(x+y)}$

$\Rightarrow \frac{nx-my}{y}\times \frac{dy}{dx}=\frac{nx-my}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{y}{x}$

Thus $y^{\prime }=\frac{y}{x}$