Question

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

given equation is $x^m\cdot y^n={(x+y)}^{m+n}$

Applying logarithm on both sides gives

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

Differentiating the equation with respect to $x$

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

now separating the terms gives

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

$⟹\left(\frac{1}{y}\right)\frac{dy}{dx}=\frac{1}{x}$

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