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

Given equation is ${\left(x+y\right)}^{m+n}=x^m{y}^n$

Apply logarithms on both sides

$\left(m+n\right)\log \left(x+y\right)=m\log x+n\log y$

Differentiate both sides

$\begin{array}{rcl}\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right)& =& \frac{m}{x}+\frac{n}{y}\frac{dy}{dx}\\ \frac{dy}{dx}\left(\frac{m+n}{x+y}-\frac{n}{y}\right)& =& \frac{m}{x}-\frac{m+n}{x+y}\\ \frac{dy}{dx}\left(\frac{my+ny-nx-ny}{y\left(x+y\right)}\right)& =& \frac{mx+my-mx-nx}{x\left(x+y\right)}\\ \frac{dy}{dx}\left(\frac{my-nx}{y\left(x+y\right)}\right)& =& \frac{my-nx}{x\left(x+y\right)}\end{array}$

Cancel out the common terms

$\frac{dy}{dx}=\boxed{\frac{\mathbf{y}}{\mathbf{x}}}$