Question

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

Answer 1

The correct answer is $⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{xe^{\sqrt{x}}}},x>0$
Let $y=\sqrt{e^{\sqrt{x}}}$
Then,$y^2=e^{\sqrt{x}}$
By differentiating this relationship with respect to $x$,we obtain
$\implies y^2=e^{\sqrt{x}}$
$⇒2y\frac{dy}{dx}=e^{\sqrt{x}}\frac{d}{dx}(\sqrt{x})$ [By applying the chain rule]
$⇒2y\frac{dy}{dx}=e^{\sqrt{x}}\frac{1}{2}.\frac{1}{\sqrt{x}}$
$⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4y\sqrt{x}}$
$⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{e^{\sqrt{x}}}\sqrt{x}}$
$⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{xe^{\sqrt{x}}}},x>0$