Question

If $xy\ne 0,x+y\ne 0$ and $x^m{y}^n=(x+y{)}^{m+n}$ where $m,n\in N$ then $\frac{dy}{dx}=$

• A. $\frac{y}{x}$
• B. $\frac{x+y}{xy}$
• C. $xy$
• D. $\frac{x}{y}$

Answer 1

Answer: (A)

$\frac{y}{x}$

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

$\log [x{]}^m+\log [y{]}^n=(m+n)\log (x+y)$

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

Differentiate

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

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

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

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

$\frac{dy}{dx}\frac{[nx-my]}{y(x+y)}=\frac{nx-my}{x(x+y)}$

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