Question

Differentiate the given function w.r.t. $x$.
$y=\sqrt{e^{\sqrt{x}}},x>0$

Answer 1

Let $y=\sqrt{e^{\sqrt{x}}},x>0$
$\Rightarrow y^2=e^{\sqrt{x}}$
Differentiating both sides w.r.t $x$
$\Rightarrow 2y\frac{dy}{dx}=e^{\sqrt{x}}\frac{d}{dx}\left(\sqrt{x}\right)$ [By applying the chain rule]

$\Rightarrow 2y\frac{dy}{dx}=e^{\sqrt{x}}\frac{1}{2}.\frac{1}{\sqrt{x}}$

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

$\Rightarrow \frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{e^{\sqrt{x}}}\sqrt{x}}$

$\Rightarrow \frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{x{e}^{\sqrt{x}}}},x>0$