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Question Number 11552 by Nayon last updated on 28/Mar/17

$$find\frac{d}{dx}\left(y\right)wherey=\stackrel{\sqrt{x}}{}\sqrt{\sqrt{x}}$$$$$$

Answered by ajfour last updated on 28/Mar/17

$$y={\left(\sqrt{x}\right)}^{\frac{1}{\sqrt{x}}}$$$$\ln y=\frac{\ln \sqrt{x}}{\sqrt{x}}$$$$\frac{1}{y}\frac{dy}{dx}=\left(\frac{\left(\frac{1}{\sqrt{x}}\right)\left(\frac{1}{2\sqrt{x}}\right)\sqrt{x}-\frac{\ln \sqrt{x}}{2\sqrt{x}}}{x}\right)$$$$\frac{dy}{dx}=y\left(\frac{1-\ln \sqrt{x}}{2x\sqrt{x}}\right)$$$$\frac{dy}{dx}=\frac{{\left(\sqrt{x}\right)}^{\frac{1}{\sqrt{x}}}(1-\ln \sqrt{x})}{2x\sqrt{x}}.$$

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