# 7. Differentiate the following w.r.t. x: $\sqrt { e ^{ \sqrt x }} , x > 0$

Given function is
$f(x)=\sqrt { e ^{ \sqrt x }}$
Lets take $g(x ) = \sqrt x$
Now, our function reduces to
$f(x) = \sqrt {e^{g(x)}}$
Now,
$f^{'}(x) = g^{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.\frac{d({e^{g(x)}})}{dx} = g{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.{e^{g(x)}} = \frac{g^{'}(x).e^{g(x)}}{2.\sqrt{e^{g(x)}}} = \frac{g^{'}(x).e^{\sqrt x}}{2.\sqrt{e^{\sqrt x}}}$                   -(i)
And
$g(x)=\sqrt x\\\Rightarrow g^{'}(x) = \frac{(\sqrt x)}{dx}=\frac{1}{2\sqrt x}$
Put this value in our equation (i)
$f^{'}(x) =\frac{e^{\sqrt x}}{2\sqrt x.2.\sqrt{e^{\sqrt x}}} = \frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}$
Therefore, the answer is $\frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}.\ \ x>0$

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