यदि $y = \sqrt{x}$, तो $\frac{d^2y}{dx^2}$ क्या है?

हल

1. 01

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

2. 02

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