Question

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

Answer 1

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

Taking log both sides we get

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

Differentiating w.r.t. x we get

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

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

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

$\Rightarrow$ $\frac{dy}{dx}=\left(\frac{nx-my}{nx-my}\right)\frac{y}{x}=\frac{y}{x}$ $\Rightarrow$ $\frac{dy}{dx}=\frac{y}{x}$