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Question Number 137280 by mathlove last updated on 31/Mar/21

Answered by EDWIN88 last updated on 31/Mar/21

⇔ y = x^(((π/(ln x)))) = x^(πlog _x (e))   ⇔ y= x^(log _x (e^π ))  ; y = e^π  ⇒(dy/dx) = 0.

$$\Leftrightarrow\:\mathrm{y}\:=\:\mathrm{x}^{\left(\frac{\pi}{\mathrm{ln}\:\mathrm{x}}\right)} =\:\mathrm{x}^{\pi\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{e}\right)} \\ $$$$\Leftrightarrow\:\mathrm{y}=\:\mathrm{x}^{\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{e}^{\pi} \right)} \:;\:\mathrm{y}\:=\:\mathrm{e}^{\pi} \:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{0}. \\ $$$$ \\ $$

Answered by mr W last updated on 31/Mar/21

y=x^(π/(ln x))   ln y=(π/(ln x))×ln x=π  ⇒y=e^π   ⇒(dy/dx)=0

$${y}={x}^{\frac{\pi}{\mathrm{ln}\:{x}}} \\ $$$$\mathrm{ln}\:{y}=\frac{\pi}{\mathrm{ln}\:{x}}×\mathrm{ln}\:{x}=\pi \\ $$$$\Rightarrow{y}={e}^{\pi} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\mathrm{0} \\ $$

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