Question

find dy/dx when x=y(1 + log x)​

Answer 1

Given,

$\longrightarrow x=y(1+\log x)$

Differentiating wrt $x,$

$\longrightarrow \dfrac{dx}{dx}=\dfrac{d}{dx}\left[y(1+\log x)\right]$

Applying product rule,

$\longrightarrow 1=\dfrac{dy}{dx}(1+\log x)+\dfrac{y}{x}$

$\longrightarrow 1=\dfrac{dy}{dx}(1+\log x)+\dfrac{y(1+\log x)}{x(1+\log x)}$

$\longrightarrow 1=\dfrac{dy}{dx}(1+\log x)+\dfrac{x}{x(1+\log x)}$

$\longrightarrow 1=\dfrac{dy}{dx}(1+\log x)+\dfrac{1}{1+\log x}$

$\longrightarrow \dfrac{dy}{dx}(1+\log x)=1-\dfrac{1}{1+\log x}$

$\longrightarrow \dfrac{dy}{dx}(1+\log x)=\dfrac{\log x}{1+\log x}$

$\longrightarrow\underline{\underline{\dfrac{dy}{dx}=\dfrac{\log x}{\left(1+\log x\right)^2}}}$