Question

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

• A. $\frac{y}{2x}$
• B. $\frac{2y}{x}$
• C. $-\frac{y}{x}$
• D. $\frac{y}{x}$

Answer 1

Answer: (D)

$\frac{y}{x}$

We have,

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

On differentiating both sides w.r.t $x$, we get

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

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

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

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

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

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

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

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

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

$\frac{1}{x}=\frac{1}{y}\frac{dy}{dx}$

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

Hence, this is the answer.