If \({\rm{s}} = \sqrt {{{\rm{t}}^2} + 1} \), then \(\frac{{{{\rm{d}}^2}{\rm{s}}}}{{{\rm{d}}{{\rm{t}}^2}}}\) is equal to

1
\(\frac{1}{{\rm{s}}}\)
2
\(\frac{1}{{{{\rm{s}}^2}}}\)
3
\(\frac{1}{{{{\rm{s}}^3}}}\)
4
\(\frac{1}{{{{\rm{s}}^4}}}\)

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