Find $\displaystyle \frac{dy}{dx}$, where $\displaystyle y = (\sqrt{x})^{x^{x^{x^{x\mbox{...}\infty}}}}$
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Originally Posted by fardeen_gen Find $\displaystyle \frac{dy}{dx}$, where $\displaystyle y = (\sqrt{x})^{x^{x^{x^{x\mbox{...}\infty}}}}$ $\displaystyle y^2 = (x)^{x^{x^{x^{x\mbox{...}\infty}}}}$ $\displaystyle \log y^2 = (x)^{x^{x^{x^{x\mbox{...}\infty}}}} \log x$ $\displaystyle \log y^2 = y^2\log x $ Can you continue?
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