If $y=7 \sqrt{x}+3 e^x$, then $\frac{d^2 y}{d x^2}$ is:

$\frac{7}{2 \sqrt{x}}+3 e^x$ $\frac{-7}{4} x^{-\frac{3}{2}}$ $\frac{-7}{4} x^{\frac{3}{2}}$ $\frac{7}{2} x^{\frac{-3}{2}}+e^x$

$\frac{7}{2 \sqrt{x}}+3 e^x$

The correct answer is Option (1) - $\frac{7}{2 \sqrt{x}}+3 e^x$