Differentiate the following w.r.t. x:


Differentiate the following w.r.t. x:

$\sqrt{e^{\sqrt{x}}}, x>0$


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

Then, $y^{2}=e^{\sqrt{x}}$

By differentiating this relationship with respect to x, we obtain


$\Rightarrow 2 y \frac{d y}{d x}=e^{\sqrt{x}} \frac{d}{d x}(\sqrt{x})$     [By applying the chain rule]

$\Rightarrow 2 y \frac{d y}{d x}=e^{\sqrt{x}} \frac{1}{2} \cdot \frac{1}{\sqrt{x}}$

$\Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 y \sqrt{x}}$

$\Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{e^{\sqrt{x}}} \sqrt{x}}$

$\Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{x e^{\sqrt{x}}}}, x>0$

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