Question Number 26300 by d.monhbayr@gmail.com last updated on 23/Dec/17
$${y}={a}^{\mathrm{arc}{tg}\sqrt{{x}}} \\ $$$${y}'=? \\ $$
Commented by abdo imad last updated on 24/Dec/17
$${we}\:{have}\:\:{y}=\:{e}^{{arctan}\left(\sqrt{\left.{x}\right)}{lna}\:\right.} \:\:\:{with}\:{condition}\:{a}>\mathrm{0} \\ $$$${dy}/{dx}={lna}\frac{{d}}{{dx}}\left(\:{arctan}\left(\sqrt{{x}}\right)\right){e}^{{srctan}\left(\sqrt{{x}}\right){lna}} \\ $$$$=\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}}{\mathrm{1}+{x}}\:{lna}.{e}^{{artan}\left(\sqrt{{x}}\right){lna}} \\ $$$$=\:\frac{{lna}}{\mathrm{2}\sqrt{{x}}\left(\mathrm{1}+{x}\right)}\:{e}^{{arctan}\left(\sqrt{\left.{x}\right)}\right.} \:^{{lna}} \:\:\:. \\ $$