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If $ y= \sqrt{e^{\sqrt x}}$ , then $\large\frac{dy}{dx}$ is

$\begin{array}{1 1} \frac{e^{1/2 \sqrt x}}{4 \sqrt x} \\ \frac{e^{-1/2 \sqrt x}}{4 \sqrt x} \\ \frac{e^{2 \sqrt x}}{ \sqrt x} \\ \frac{e^{2 \sqrt x}}{4 \sqrt x}\end{array} $

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Answer :
$\frac{e^{1/2 \sqrt x}}{4 \sqrt x}$
$\sqrt x =t=>e^{\sqrt x} =e^t =u$
$y= \sqrt {u}$
$\large\frac{dy}{du} =\frac{1}{2} u^{-1/2}$
$\therefore \large\frac{dy}{dx} =\frac{dy}{du} \times \frac{du}{dt} \times \frac{dt}{dx}$
$\qquad= \large\frac{1}{2 \sqrt u} $$ \times e^t \times \frac{1}{2 \sqrt x} $
Answer :$\frac{e^{1/2 \sqrt x}}{4 \sqrt x}$
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