Answer
$\frac{dy}{dx} = 2e^{(4 \sqrt x + x^{2})} \times (\frac{1}{\sqrt x} + x)$
Work Step by Step
$y = e^{u}$
where,
$u = 4 \sqrt x + x^{2}$
Now,
$\frac{dy}{du} = e^{u}$
and,
$\frac{du}{dx} = \frac{2}{\sqrt x} + 2x$
So,
$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$
$\frac{dy}{dx} = e^{u} \times (\frac{2}{\sqrt x} + 2x)$
$\frac{dy}{dx} = 2e^{(4 \sqrt x + x^{2})} \times (\frac{1}{\sqrt x} + x)$
