# If $x^my^n=(x+y)^{m+n}$,Prove that $(i)\;\large\frac{dy}{dx}=\frac{y}{x}$

This is the first part of the multi-part question Q80

Toolbox:
• A function $f(x,y)$ of two variables $x$ and $y$,is said to be implicit,which are jumbled in such a way,that it is not possible to write y exclusively as a function of $x$.
• $\large\frac{d}{dx}$$\phi(y)=\large\frac{d}{dy}.$$\phi(y).\large\frac{dy}{dx}$
Step 1:
Given : $x^{\large m}y^{\large n}=(x+y)^{\large m+n}$
Take $\log$ on both sides
$\log x^m+\log y^n=(m+n)\log (x+y)$
$\Rightarrow m\log x+n\log y=(m+n)\log (x+y)$
Differentiating w.r.t $x$ on both sides we get,
$m.\large\frac{1}{x}+$$n\large\frac{1}{y}\frac{dy}{dx}=$$(m+n).\large\frac{1}{(x+y)}$$\big(1+\large\frac{dy}{dx}\big) \Rightarrow \large\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{(m+n)}{(x+y)}$$\big(1+\large\frac{dy}{dx}\big)$
Step 2:
$\large\frac{dy}{dx}\big(\large\frac{n}{y}-\frac{(m+n)}{(x+y)}\big)=\large\frac{(m+n)}{(x+y)}-\frac{m}{x}$
$\Rightarrow \large\frac{dy}{dx}\bigg(\large\frac{n(x+y)-y(m+n)}{y(x+y)}\bigg)=\large\frac{x(m+n)-m(x+y)}{x(x+y)}$
On simplifying we get,
$\Rightarrow \large\frac{dy}{dx}\big(\large\frac{nx-my}{y}\big)=\large\frac{nx-my}{x}$
Therefore $\large\frac{dy}{dx}=\frac{y}{x}$
Hence proved.