Answer
${dy}=e^{(\tan^{-1}{\sqrt{(x^2+1)}})}\times{\frac{x}{(2+x^2)(\sqrt{(1+x^2)})}}{dx}$
Work Step by Step
We are given: $y=e^{(\tan^{-1}{\sqrt{(x^2+1)}})}$
on differentiating both sides:
${\frac{dy}{dx}}={\frac{d(e^{(\tan^{-1}{\sqrt{(x^2+1)}}})}{dx}}$
on applying the chain rule:
${\frac{dy}{dx}}=e^{(\tan^{-1}{\sqrt{(x^2+1)}})}\times{\frac{1}{(1+(\sqrt{(1+x^2)})^2)}}{\frac{d({({\sqrt{(x^2+1)}})})}{dx}}$
${\frac{dy}{dx}}=e^{(\tan^{-1}{\sqrt{(x^2+1)}})}\times{\frac{1}{(2+x^2)}}{\frac{2x}{2\sqrt{(x^2+1)}}}$
${dy}=e^{(\tan^{-1}{\sqrt{(x^2+1)}})}\times{\frac{x}{(2+x^2)(\sqrt{(1+x^2)})}}{dx}$
The final answer is: ${dy}=e^{(\tan^{-1}{\sqrt{(x^2+1)}})}\times{\frac{x}{(2+x^2)(\sqrt{(1+x^2)})}}{dx}$