Question

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

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

Answer 1

Answer: (D)

$\because x^{m} \cdot y^{m} =\left(x+y\right)^{m+n}$
Taking log on both sides, we get
$m\, log\, x+n\, log\, y =\left(m+n\right) log \,\left(x+y\right)$
On differentiating with respect to $x$, we get
$\frac{m}{x}+\frac{n}{y} \frac{d y}{d x}=\frac{\left(m+n\right)}{\left(x+y\right)} \left(1+\frac{d y}{d x}\right)$
$\Rightarrow \, \frac{d y}{d x} \left(\frac{m+n}{x+y}-\frac{n}{y}\right)=\frac{m}{x}-\frac{m+n}{x+y}$
$\Rightarrow \, \frac{d y}{d x} \left(\frac{my+ny-nx-ny}{y \left(x+y\right)}\right)$
$=\frac{m x +my-mx-nx}{x\left(x+y\right)}$
$\Rightarrow \, \frac{d y}{d x}=\frac{y}{x}$