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I totally have no idea about this...
If $\frac{dy}{dx} = \sqrt{y^2+1}$, then $\frac{d^2y}{dx^2}$=?
The correct answer is $y$.

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up vote 4 down vote accepted

It's just a few applications of the chain rule and use of the ODE as $\frac d{dx} y(x)$ appears. $$\begin{align*} \frac d{dx} \sqrt{y^2(x)+1} & = \frac1{2\sqrt{y^2(x)+1}} \cdot \frac d{dx} (y^2(x) + 1) \\ & = \frac1{2\sqrt{y^2(x)+1}} 2y(x) \cdot y'(x) \\ & \stackrel{\text{ODE}}= \frac1{\sqrt{y^2(x)+1}} \cdot y(x) \cdot \sqrt{y^2(x)+1} \\ & = y(x) \end{align*}$$

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Hint: $\frac{d^{2}y}{dx^2} = \left( \frac{dy}{dx}\right)'$

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Hint: Use chain rule while differentiating $\frac{dy}{dx}$ with respect to $x$.

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