Question

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

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

Answer 1

The correct option is A $\frac{y}{x}$

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

Taking log both sides

$log\left(x^m{y}^n\right)=log(x+y{)}^{m+n}$

or, $mlogx+nlogy=(m+n)log(x+y)$

Differentiating both sides w.r.t. x-

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

or, $\frac{m}{n}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}+\frac{m+n}{x+y}\frac{dy}{dx}$

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

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

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

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

or, $\frac{dy}{dx}=\frac{y}{x}$ .........$\left(1\right)$

Again differentiating both sides wrt x-

$\frac{d^{2}y}{d{x}^2}=\frac{x\frac{dy}{dx}-y.1}{x^2}$

or, $\frac{d^{2}y}{d{x}^2}=\frac{x\frac{y}{x}-y}{x^2}$ [Using $\left(1\right)$]

or, $\frac{d^{2}y}{d{x}^2}=\frac{y-y}{x^2}$

or, $\frac{d^{2}y}{d{x}^2}=0$.