hope you got it I did my best

Question:

[tex] \sf \: If \: {x}^{y} = {e}^{x - y} , \: show \: that \: \dfrac{dy}{dx} = \dfrac{logx}{ {(1 + logx)}^{2} } \\ [/tex]

Explanation:

Given that,

[tex] \sf \: {x}^{y} = {e}^{x - y} \\ [/tex]

On taking log on both sides, we get

[tex] \sf \: log ({x}^{y}) = log( {e}^{x - y} ) \\ [/tex]

[tex] \sf \: ylog x = (x - y)loge \\ [/tex]

[tex] \sf \: ylog x = x - y \\ [/tex]

[tex] \sf \: ylog x + y = x \\ [/tex]

[tex] \sf \: y(log x + 1) = x \\ [/tex]

[tex] \sf \: y = \dfrac{x}{1 + logx} \\ [/tex]

On differentiating both sides w. r. t. x, we get

[tex] \sf \: \dfrac{d}{dx}y = \dfrac{d}{dx} \bigg( \dfrac{x}{1 + logx} \bigg) \\ [/tex]

[tex] \sf \: \dfrac{dy}{dx} = \dfrac{(1 + logx)\dfrac{d}{dx}(x) - x\dfrac{d}{dx}(1 + logx)}{ {(1 + logx)}^{2} } \\ [/tex]

[tex] \sf \: \dfrac{dy}{dx} = \dfrac{(1 + logx) (1)- x \times \dfrac{1}{x} }{ {(1 + logx)}^{2} } \\ [/tex]

[tex] \sf \: \dfrac{dy}{dx} = \dfrac{1 + logx- 1 }{ {(1 + logx)}^{2} } \\ [/tex]

[tex]\implies\sf\:\dfrac{dy}{dx} = \dfrac{logx}{ {(1 + logx)}^{2} } \\ [/tex]

Hence,

[tex]\implies\sf\:\boxed{\bf\:\dfrac{dy}{dx} = \dfrac{logx}{ {(1 + logx)}^{2} } \: } \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Formulae Used:

[tex] \sf \: loge = 1 \\ [/tex]

[tex] \sf \: log( {x}^{y} ) = ylogx \\ [/tex]

[tex] \sf \: \dfrac{d}{dx} \dfrac{u}{v} = \dfrac{v\dfrac{d}{dx}u - u\dfrac{d}{dx}v}{ {v}^{2} } \\ [/tex]

[tex] \sf \: \dfrac{d}{dx}logx = \dfrac{1}{x} \\ [/tex]

[tex] \sf \: \dfrac{d}{dx}x = 1 \\ [/tex]

[tex] \sf \: \dfrac{d}{dx}k = 0 [/tex]