Question

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

• A. $\frac{x+y}{xy}$
• B. $xy$
• C. $0$
• D. $\frac{y}{x}$

Answer 1

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

Consider $x^m{y}^n={(x+y)}^{m+n}$ and take log both sides:

$\ln \left(x^m{y}^n\right)=\ln {(x+y)}^{m+n}\Rightarrow \ln x^m+\ln y^n=\ln {(x+y)}^{m+n}(\because \ln \left(ab\right)=\ln a+\ln b)\Rightarrow m\ln x+n\ln y=(m+n)\ln (x+y)(\because \ln a^x=x\ln a)$

Differentiate both sides with respect to $x$

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

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