If $y = \sqrt{x}$, what is $\frac{d^2y}{dx^2}$?

Explanation

1. 01

   $y = x^{1/2}$, so $\frac{dy}{dx} = \frac{1}{2}x^{-1/2}$.

2. 02

   $\frac{d^2y}{dx^2} = \frac{1}{2} \cdot \left(-\frac{1}{2}\right)x^{-3/2} = -\frac{1}{4x^{3/2}}$.