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]
\(⇒2y\frac{dy}{dx}=e^{\sqrt{x}}\frac{1}{2}.\frac{1}{\sqrt{x}}\)
\(⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4y\sqrt{x}}\)
\(⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{e^{\sqrt{x}}}\sqrt{x}}\)
\(⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{xe^{\sqrt{x}}}},x>0\)