Question

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

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

Answer 1

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

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