<s>[INST] find the value of x at second derivative The quesion is :
\begin{align}
   \ y= \sqrt{x} ~\text{ find the value of x for which}~ y+ \frac{d^2y}{dx^2} = 0
    \end{align}
\begin{align}
   \ 
    \end{align}
The answer given is :   x= 1/2
I am not sure how they got this answer so any help would be appreciated.
 [/INST] $y=\sqrt { x } $ so plug this  in the equation $$\sqrt { x } +\frac { d }{ dx } \left( \frac { d\left( \sqrt { x }  \right)  }{ dx }  \right) =0\\$$ now here $\frac { d }{ dx } \left( \frac { d\left( \sqrt { x }  \right)  }{ dx }  \right) $ means you should find derivative twice of $y=\sqrt { x } $
$$ \sqrt { x } +\frac { d }{ dx } \left( \frac { 1 }{ 2\sqrt { x }  }  \right) =0\\ \sqrt { x } -\frac { 1 }{ 4x\sqrt { x }  } =0\\ 4{ x }^{ 2 }-1=0\\ x=1/2$$
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