Continuity and Differentiability — Class 12 Maths Solution

Let $y = \sqrt {{e^{\sqrt x }}}$

therefore, $\cfrac{{dy}}{{dx}} = \cfrac{d}{{dx}}\left( {\sqrt {{e^{\sqrt x }}} } \right) = \cfrac{d}{{dx}}{({e^{\sqrt x }})^{1/2}} = \cfrac{1}{2}{({e^{\sqrt x }})^{ - \cfrac{1}{2}}} \cdot \cfrac{d}{{dx}}{e^{\sqrt x }}$

$= \cfrac{1}{2}{({e^{\sqrt x }})^{ - \cfrac{1}{2}}} \cdot {e^{\sqrt x }}\cfrac{d}{{dx}}\sqrt x = \cfrac{1}{2}{({e^{\sqrt x }})^{ - 1/2}} \cdot {e^{\sqrt x }} \cdot \cfrac{1}{2}{(x)^{ - 1/2}}$

$= \cfrac{{{e^{\sqrt x }}}}{{4\sqrt {{e^x}} \sqrt x }},x > 0$