If ylogx=(x-y) ,prove that dy/dx=logx/(1+logx)^2

{tex}\text {Given equation is }y \text { log }x = x-y{/tex}            {tex}...(1){/tex}

{tex}\therefore x=y \text { log }x + y{/tex}

{tex}\Rightarrow x=y(\text {log }x+1)=y(1 + \text {log } x){/tex}          {tex}...(2){/tex}

{tex}\text {Differentiating Eqn. (1) using product rule, w.r.t. }x, \text { we get,}{/tex}

{tex}\cfrac {y}{x} + \text {log } x.\cfrac {dy}{dx} = 1-\cfrac {dy}{dx}{/tex}

{tex}\Rightarrow \text {log }x.\cfrac {dy}{dx} + \cfrac {dy}{dx} =1-\cfrac {y}{x}{/tex}

{tex}\Rightarrow \cfrac {dy}{dx} \bigg(\text {log }x+1\bigg)=\cfrac {x-y}{x}{/tex}

{tex}\Rightarrow \cfrac {dy}{dx}=\cfrac {x-y}{x(1+\text {log }x)}{/tex}

{tex}\text {From Eqn. (1) and Eqn. (2),}{/tex}

{tex}\Rightarrow \cfrac {dy}{dx} =\cfrac {y \text { log }x}{y(1+\text { log }x)(1+\text {log }x)}{/tex}

{tex}\Rightarrow \cfrac{dy}{dx}=\cfrac {\text { log }x}{(1+\text {log }x)^2}{/tex}{tex}\text {Proved}{/tex}