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Question Number 26368 by d.monhbayr@gmail.com last updated on 24/Dec/17

$$y=a^{\mathrm{arc}tg\sqrt{x}}$$$$derivative?$$

Commented by kaivan.ahmadi last updated on 24/Dec/17

$${\log }_{\mathrm{a}}\mathrm{y}=\mathrm{arctg}\sqrt{\mathrm{x}}\Rightarrow \frac{{\mathrm{y}}^{\prime }}{\mathrm{y}}\mathrm{lna}=\frac{{\left(\sqrt{\mathrm{x}}\right)}^{{}^{\prime }}}{1+{\left(\sqrt{\mathrm{x}}\right)}^2}=$$$$\frac{\frac{1}{2\sqrt{\mathrm{x}}}}{1+\mathrm{x}}=\frac{1}{2\sqrt{\mathrm{x}}(1+\mathrm{x})}\Rightarrow {\mathrm{y}}^{{}^{\prime }}=\frac{\mathrm{y}}{2\mathrm{lna}\sqrt{\mathrm{x}}(1+\mathrm{x})}\Rightarrow$$$${\mathrm{y}}^{{}^{\prime }}=\frac{{\mathrm{a}}^{\mathrm{arctg}\sqrt{\mathrm{x}}}}{2\mathrm{lna}\sqrt{\mathrm{x}}(1+\mathrm{x})}$$

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