Question

If $X^m.Y^n={(X+Y)}^{m+n}$ then prove that $\frac{dy}{dx}=\frac{y}{x}$

Answer 1

$x^m{y}^n=(x+y{)}^{m+n}takelogbothsideslog(x^m{y}^n)=log(x+y{)}^{m+n}log{x}^m+log{y}^n=(m+n)log(x+y)mlogx+nlogy=(m+n)log(x+y)nowdifferentiatebothsidesw.r.t"x"m\left(\frac{dlogx}{dx}\right)+n\left(\frac{dlogy}{dx}\right)=(m+n)\left(\frac{dlog(x+y)}{dx}\right)\left(\frac{m}{x}\right)+n\left(\frac{1}{y}\right)\times \left(\frac{dy}{dx}\right)=(m+n)\times \left(\frac{1}{(x+y)}\right)(1+(\frac{dy}{dx}\left)\right)\left(\frac{m}{x}\right)+\left(\frac{n}{y}\right)\left(\frac{dy}{dx}\right)=\left(\frac{m+n}{x+y}\right)+\left(\frac{m+n}{x+y}\right)\left(\frac{dy}{dx}\right)$

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