Question

If x^myⁿ = (x + y)^m+n, prove that \[\frac{dy}{dx} = \frac{y}{x} .\]

Answer 1

Given: x^myⁿ = (x + y)^m+n

​Taking log on both the sides, we get

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

\[ \Rightarrow \log\left( x^m \right) + \log\left( y^n \right) = \left( m + n \right) \log\left( x + y \right)\]

\[ \Rightarrow m\log x + n\log y = \left( m + n \right) \log\left( x + y \right)\]

Differentiating w.r.t. x, we get

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

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

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

\[ \Rightarrow \frac{dy}{dx}\left( \frac{my - nx}{y} \right) = \left( \frac{my - nx}{x} \right)\]

\[ \therefore \frac{dy}{dx} = \frac{y}{x}\]

Hence proved.