Question
Differentiate the $\sqrt{e^{\sqrt{x}}}, x>0$ w.r.t. x.

Answer

Let $y=\sqrt{e^{\sqrt{x}}}$ 
Then, $y^{2}=e^{\sqrt{x}}$ 
Now, differentiating both sides w.r.t x, we get,
$2 y \frac{d y}{d x}=e^{\sqrt{x}} \frac{d}{d x}(\sqrt{x})$ 
$=e^{\sqrt{x}} \frac{1}{2} \cdot \frac{1}{\sqrt{x}}$ 
$\Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 y \sqrt{x}}$ 
$\Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{e^{\sqrt{x}}} \sqrt{x}}$ 
$\Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{x e^{\sqrt{x}}}}$

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