Question

Find $\frac{dy}{dx}$ for $(x+y{)}^{m+n}=x^m{y}^n$.

Answer 1

Given,

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

taking $\log$ on both sides, we get,

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

differentiating on both sides, we get,

$\frac{m+n}{x+y}(1+y^{\prime })=\frac{m}{x}+\frac{n}{y}{y}^{\prime }$

$\frac{m+n}{x+y}+\frac{m+n}{x+y}{y}^{\prime }=\frac{m}{x}+\frac{n}{y}{y}^{\prime }$

$\frac{m+n}{x+y}{y}^{\prime }-\frac{n}{y}{y}^{\prime }=\frac{m}{x}-\frac{m+n}{x+y}$

$y^{\prime }(\frac{m+n}{x+y}-\frac{n}{y})=\frac{m}{x}-\frac{m+n}{x+y}$

$y^{\prime }\left(\frac{my-nx}{(x+y)y}\right)=\frac{my-nx}{(x+y)x}$

$\therefore y^{\prime }=\frac{dy}{dx}=\frac{y}{x}$