If $\frac{d y}{d x} = \sqrt{y^{2} + 1}$ $(dy)/(dx)=sqrt(y^2+1)$ , then $\frac{d^{2} y}{d^{2} x}$ $(d^2y)/(d^2x)$ is what?

Question

I took the derivative of this function and got $\frac{y}{\sqrt{y^{2} + 1}}$ $y/(sqrt(y^2+1))$ . However, the correct answer is $\sqrt{2 y}$ $sqrt(2y)$ . How?

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Answer 1 · 2018-05-01T06:09:01

$\frac{d^{2} y}{{d x}^{2}} = y$ $(d^2y)/(dx^2)=y$
(not $\sqrt{2} y$ $sqrt2y$ )

As $\frac{d y}{d x} = \sqrt{y^{2} + 1}$ $(dy)/(dx)=sqrt(y^2+1)$

$\frac{d^{2} y}{{d x}^{2}} = \frac{1}{2 \sqrt{y^{2} + 1}} \cdot 2 y \cdot \frac{d y}{d x}$ $(d^2y)/(dx^2)=1/(2sqrt(y^2+1)) * 2y * (dy)/(dx)$

= $\frac{1}{2 \sqrt{y^{2} + 1}} \cdot 2 y \cdot \sqrt{y^{2} + 1}$ $1/(2sqrt(y^2+1)) * 2y * sqrt(y^2+1)$

= $y$ $y$

Answer 2 · 2018-05-01T06:23:57

$\Rightarrow \frac{d^{2} y}{{d x}^{2}} = y$ $=> (d^2y)/(dx^2) = y$

$\Rightarrow {(\frac{d y}{d x})}^{2} = y^{2} + 1$ $=> ((dy)/(dx))^2 = y^2 + 1$

$\Rightarrow 2 \frac{d y}{d x} \frac{d^{2} y}{{d x}^{2}} = 2 y \frac{d y}{d x}$ $=> 2 (dy)/(dx) (d^2y)/(dx^2) = 2y (dy)/(dx)$

$=> cancel(2 (dy)/(dx)) (d^2y)/(dx^2) = cancel(2 (dy)/(dx))y$

$=> (d^2y)/(dx^2) = y$