Answer
$g'(u)=ue^{\sqrt secu^{2}}(\sqrt secu^{2})(tanu^{2})$
Work Step by Step
$g'(u)=\frac{d}{du}(e^{\sqrt secu^{2}})$
$=e^{\sqrt secu^{2}}\frac{d}{du}({\sqrt secu^{2}})$
$=e^{\sqrt secu^{2}}\frac{1}{2(\sqrt secu^{2})}\frac{d}{du}({secu^{2}})$
$=e^{\sqrt secu^{2}}\frac{1}{2(\sqrt secu^{2})}({secu^{2}})({tanu^{2}}).2u$
Hence, $g'(u)=ue^{\sqrt secu^{2}}(\sqrt secu^{2})(tanu^{2})$
